projectile motion going to be the topic of this lesson in my new General Physics playlist now if you find yourself struggling with this you're in good company because this is a topic that a good chunk of students struggle with so but we're going to break it down we're going to find out that it really is just kind of a synthesis of two one-dimensional problems combined into a single problem and that we really can treat the X Dimension and Y Dimension independently as we as was demonstrated in the last lesson my name is Chad and welcome to Chad's prep where my goal is to take the stress out of learning science now if you're new to the channel we've got comprehensive playlists for General chemistry organic chemistry General Physics and high school chemistry and on chatsprep.com you'll find premium Master courses for the same that include study guides and a ton of practice you'll also find comprehensive prep courses for the DAT the MCAT and the oat all right so let's introduce projectile motion a little bit and uh we're going to talk about it this is our introduction to two-dimensional problems now the first two problems we're going to do are going to be one-dimensional problems instead of two-dimensional problems but they're going to set us up for the first actual 2D problem so the idea is that the initial velocity is going to have oftentimes both X and Y components so we're going to treat those different components those different dimensions completely independent of one another so if you take a look at what's going on in the X Dimension as long as we're ignoring air resistance there's no drag on an object's Motion in this case and so it's going to have constant velocity well if it's constant velocity that means no acceleration and we have just a single equation therefore to deal with that motion in the X Dimension but in the Y Dimension there is gravity and as long as you're near the surface of the Earth that gravity has a constant magnitude of negative 9.8 meters per second squared and that negative here again refers to the fact that it points downward and that is constant everywhere along its Journey that acceleration has exactly that value as long as the entire time it's near the surface of the Earth which is customary for us to deal with in this section all right so we take a look at the motion here there's a couple other things I want you to know is so we're going to break up its initial velocity into components the next component a y component uh same way we've learned to split vectors up into components and things of this sort but big thing to note is that uh the x velocity is never going to change because there's nothing speeding up or slowing it down again in that horizontal Dimension but in the vertical Dimension that velocity is going to change because the acceleration due to gravity operates in the vertical Direction pointing downward and so on the way up it's going to slow down the Y component of the Velocity all the way till it reaches that Max height when it reaches that Max height that y component of the velocity is going to be zero so just like when you throw a marker straight up in the air so at some point it's going to reach its Max height and change directions well all the way up its velocity has been positive but it's been getting slower and slower and slower at the top instantaneously it has velocity of zero before it heads back down and its velocity gets more and more negative all the way down faster and faster and faster in the negative Direction cool same thing works here so it's still going to have an X component to its velocity up there so its overall velocity is not necessarily zero in this case but the Y component will go to zero at that maximum height so and then all this uh incidentally oddly enough gravity is going to have just as much time to operate it on the way back down and so right at the end here the final y velocity is going to be equal to whatever the original was assuming it is hitting the ground at the same level from which it was launched now this won't be true for all problems you'll have some where it's going to be launched off a building and then hit the ground far lower well then that won't be true so but as long as it's being launched and landing at exactly the same height I.E like ground level in this case yeah then that final y velocity will be equal to the initial y component of the Velocity as well same thing as you throw it up in the air so we saw as you throw it up in the air as it comes back down it'll have the same final velocity as the initial velocity on the way up if it weren't for air resistance anyways okay so let's deal with a couple of problems here so and again you're going to look at this first one and be like um this is a two-dimensional projectile motion lesson Chad yes and we're going to start with a couple of one-dimensional problems first though so the first one is we're gonna have a hockey puck and that hockey puck is going to be struck in the horizontal Direction sliding across the ice that uh with a constant velocity of 86.6 meters per second and I'll write that as an X component to its velocity but in this case it only has x velocity and it's just going to slide across the ice here in this case with no friction and the question is well how far does it travel in 10 seconds 10.0 seconds okay well again in this case we have no acceleration and so the only way we have calculating the displacement over the course of time is one equation that's what we're going to use here and specifically instead of using Delta R we're going to use just the X component here I realize again it's all X so but it's going to be X component of the Velocity times time and so in this case that X component of velocity is all the velocity so 86.6 meters per second times 10.0 seconds so and I picked this example for a couple reasons one it's going to set us up for our first two dimensional problem here in a little bit and two we can do the math in our head here because multiplying by 10 is just moving the decimal over one place and so our displacement here is 866 meters cool in this case 86.6 meters per second has three sig figs 10.0 has three sig figs so we want an answer in three sig figs and it's 866. there's no rounding here using our exact numbers gives us exactly 866 which has three sig figs so there we go cool and again this is just a one-dimensional problem with no acceleration constant velocity and we've been doing this for a whole whole chapter now so but again this is going to be the basis for our first two dimensional problem which is not the next one but the one after that all right so this next question is going to deal with a baseball being thrown perfectly vertically up in the air question says a baseball is thrown straight upward in the air with an initial velocity of 50.0 meters per second ignore air resistance and then there's a two-part question first is how long is the ball in the air and the second is what is the ball's maximum height so those are the two questions we want to answer here so a couple things we should realize is if it balls thrown perfectly vertical it's going to go up and then come right back down so a couple things we should realize at the maximum trajectory here it's y velocity is going to be zero in this case it's entire velocity is going to be zero because there's only vertical motion here in this case the other thing we should realize is that it's final velocity just as it's hitting the ground is is going to have the same magnitude as its initial velocity in the y direction on the way up and so in this case we might say that it's negative 50.0 meters per second if we cared so or just 50 you know 50 meters per second pointed downwards so just some things to realize right from the get-go okay so from here two parts of that question again how long is the ball in the air and then what is the maximum height so for how long so what you want to do is really split this into in half in that vertical Dimension and talk about the time up and the time down because it's the same time the time it takes to go from zero to I'm sorry from 50 to zero is going to be the same amount of time it takes to go from zero to negative 50 as well the change in velocity over the change in time equals acceleration well the acceleration hasn't changed and the change in velocity on the way up and the change in the velocity on the way down have the same magnitude so the time is going to have to be the same as well and we can kind of ballpark this so notice gravity acceleration due to gravity is negative 9.8 meters per second per second and again this is really approximately negative 10 meters per second per second and I love being able to Ballpark that our head it's really convenient this came out really close to 10 because we can do a lot of ballparking in our head so if you notice if we start off with a velocity of 50 meters per second well the velocity initially points up the acceleration points down and when velocity acceleration point in opposite directions that's an object that's slowing down and so if it starts at 50 well then one second later it's going to be slower by 9.8 meters per second well again I'm going to say that it's going to be roughly slower by about 10 meters per second and so instead of 50 meters per second after one second it's going to be down to 40 meters per second after two seconds it's going to be down to 30 meters per second for a velocity after three seconds down to 20 meters per second for the Y velocity so how long does it take to get to zero and you can see oh well if it's 10 meters per second per second that's going to take roughly five seconds now technically we're using this a quick equation without realizing we're using equation if you have an intuitive feel for what acceleration is and the numbers are nice like they are here when we ballpark gravity as as negative 10 meters per second per second so but we're really just using this equation the final velocity for the first half of the journey is zero the initial velocity was 50 meters per second gravity we were ballparking around 10 and if you realize you still take a zero I'm sorry yeah take zero minus the 50 you get negative 50 divided by the negative 10 gets you to 5 seconds so we're using an equation but if you don't have to use an equation because you have that intuitive understanding of X what acceleration is so much the better but at the end of the day if you want to use the equation use the equation so and that's what we're going to do to actually solve this here so we're going to again say V final in the Y component equals V initial y component plus a times T so we'll subtract this over again that final for just the first half of the journey 0 minus that 50 meters per second is equal to negative 10 meters per second per second times time we'll divide through we'll have negative 50 over negative 10. and we'll see that yep time would be five seconds well again let's get the exact answer let's put the exact 9.8 in there now and let our calculator do some work for us and we see that because we're going to be dividing by a little bit smaller number we're going to get a little bit longer time and so negative 50 divided by negative 9.8 it's going to get us 5.102 so I guess in this case we want uh three sig figs so 5.10 seconds but keep in mind that's just the way up that's 5.10 seconds on the way up which means on the way down it's going to be an additional 5.10 seconds and so the total time for the whole journey is actually going to be 10.2 seconds all right now that we got the total time so which is the first question the second was what is the ball's maximum height well the maximum height is really just the displacement vertically on the way up so from the height of zero to the max height that's the displacement going up and it's equal in magnitude to the displacement on the way down although on the way down that displacement would be negative but in magnitude same same value so you can just say well what's the displacement on the way up okay great how do we calculate displacement when we have uniform acceleration well we've got three different ways we've got this one this one and this one that'll have the displacement in there as long as we're just dealing with our y components and again my favorite is that first one if I know I can calculate the average velocity because I know the initial and final I go that way every time so and that's the case here our initial velocity again is 50 meters per second our final velocity for that first half of the journey is zero which means the average the average of 50 and zeros 25 meters per second and the journey is roughly five seconds 25 meters per second for five seconds 25 times five it'd be roughly 125 meters and again we can ballpark this in our head and then do the actual calculation and so in this case Delta R I'm sorry Delta y dealing in the Y Dimension equals the average the Y component the average velocity times time which again the average velocity was 25 meters per second first half of the journey is 5.10 seconds and let our calculator do a little more work for us we're gonna get 127.55 so I'll make that 127.6 meters and I carried one extra Sig fig so I can see if I need to round this because if you look at our input numbers they're both three sig figs 50.0 meters so in this case we're going to round this up to 128. cool and that is this placement going up now that's not the only way we could have done this we could have used this equation as well but we would have had a problem with this at least in one sense so if we were to try to use that second equation we said well the initial velocity is 50 meters per second plug it in there the acceleration is negative 10 I'm sorry negative 9.8 meters per second per second and we'd had time and time squared it would have been a quadratic and that would have sucked definitely not the way to go but you could have cheated a little bit and just said well but it's also equal to the displacement on the way down at least in magnitude and you could have figured out that way and that's much easier because what's the initial velocity on the way down well zero and if the initial velocity on the way down is zero you plug that in and that first term goes away and all of a sudden you don't have a quadratic you'll just be taking the square root instead and you could have figured it out that way or you could have used this one as well and again whether it's the way up or the way down you know the final velocity you know the initial velocity you know the acceleration and you could have calculated the displacement there again my preference is if I can use this first one I almost always do it's not always possible or it's not always the most convenient but if it is about a way of knowing that average velocity because I know the initial and final definitely the way I choose to go so we're finally dealing with a two-dimensional problem now and this one we're going to have a baseball hit uh let's read it a baseball is hit with an initial velocity of 100.0 meters per second at a launch angle of 30.0 degrees above the horizontal uh assume it is hit initially from ground level and that there is no air resistance three-part question how long is the ball in the air what is the maximum height reached by the ball how far away from the batter does the ball land so those are the questions we want to answer here and so they've got to realize that now so our angle here is 30 degrees so above the horizontal initial velocity again was 100 miles or 100 miles per hour 100 meters per second and now we got X and Y components it's not perfectly horizontal it's not perfectly vertical it's somewhere in between and so our X component here is going to be 100 meters per second times cosine of 30 degrees so in cosine of 30 is 0.866 so you work this out this is going to come out to 86.6 meters per second so and that should sound a little bit familiar because that was the x velocity we used back in the first one dimensional problem we did just a little bit ago now if we take a look at the Y component of the velocity so in this case it's going to equal 100 meters per second times the sine of 30 degrees and again I like choosing 30 degrees because at least one of the parts of math is going to be easy sine of 30 is one half exactly one-half and a half of a hundred is fifty meters per second which was the initial velocity of the baseball we just threw up in the air and the last problem and this is really a combination of the last two problems we've done and the first problem with the hockey puck we had an x velocity of 86.6 meters per second well in the X Dimension there's no acceleration gravity only operates in the vertical Dimension not in the horizontal and so it was constantly going to be 86.6 meters per second and the only equation we had to work with to calculate displacement was this guy that's not going to be any different now now in the second problem we did the very last one we just did the initial y velocity throwing the baseball up in the air was 50 meters per second and we figured out it was going to be 5.1 seconds on the way up in 5.2 seconds on the way back down and that's not going to change here either so the Y as far as the Y part of its concerned is exactly the same problem we just did now we've got to put the two problems together so but again in the X Dimension there's no acceleration only one equation in the Y Dimension we have gravity constantly acting downward and we've got a suite of equations of how to deal with uniform acceleration that's the deal now where do you start with on one of these problems you almost always are going to start with calculating how long how long is the Journey of even if it's not the first question asked is usually the first one you need to answer and the question you really need to answer is which part of the motion the X part of the motion or the Y part of motion is the reason that the journey ends because this ball is going to get launched and it's going to land right there now did it land because of what's going on in the X direction or because of what's going on in the y direction well if it's landing on the ground it's because of what's going on in the vertical Dimension so gravity is pulling it back down to hit the earth and so then it's going to be the Y set of you know dealing with the Y motion that's going to allow you to calculate for time now imagine for a minute that let's say let's say this represents a building here and there was a building sitting right here and the ball hits the building well now all of a sudden it's hitting the building so due to what's going on the horizontal Direction and you'd actually have to use the horizontal Dimensions motion to figure out the time in that case and we'll do one of those examples in a little bit as well but in this one it's going to hit the ground and we're going to use the Y Dimension motion to figure out the time and that's what we'll start all right and so in this case uh to figure out time we've got a handful of equations there but it's exactly the problem we just did a minute ago and in this case again initial y velocity is 50 meters per second well after one second that would be down to roughly 40 meters per second with gravity being roughly 10 meters per second per second and in this case on the way up it's slowing down on the way down it's speeding it back up so from 50 meter second to down to roughly 40 meter second to 30 meters a second to 20 meter second to 10 meter second and then finally at Max height reaching a y velocity of zero that's going to take roughly five seconds on the way up and then five more seconds on the way back down also the initial y velocity is 50 meter second we should keep in mind that it's going to be roughly not roughly it'll be exactly 50 meters per second right there as well because we're ignoring air resistance all right so but the math is exactly what we did a little bit ago and we'll say that the final y velocity equals the initial y velocity plus a t so final y velocity for the first half of the journey is zero so the initial y velocity y component of the Velocity is 50 meters per second the Y acceleration is negative 9.8 meters per second per second or meters per second squared and then times time we'll solve for time it'll get negative 50 over negative 9.8 and we already know this is going to come up to 5.10 seconds cool and that means that the total time for the way up and the way back down is just double that at 10.2 seconds again exactly the same numbers from the last problem which is why it's working out to be exactly the same amount of time okay second question is what is the max height well it's exactly what it was in the last question as well and we're going to use that time to get to the max height is 5.1 seconds and we can again we can use any one of these three equations so in in principle to figure that out these two have time this one doesn't but again we also know the initial and final y components of the velocity and the Y acceleration we could calculate using that one as well but again my favorite is definitely that first one definitely my go-to in this case so Delta Y is equal to the Y component the average velocity times time so and the average y velocity again on the way up starts with an initial of 50 meters per second to zero the average is going to be 25 therefore times that 5.1 seconds this is probably still in my calculator but I can't seem to pull it off up my Head 120 we round it to 128 right 127.55 or 127.6 we round it up to 128 meters for the max height total Journey 10.2 seconds and then the question is how far away is the ball land away from the batter horizontally sometimes we call that the range in projectile motion that X displacement and that's what we're solving for x displacement and again with no acceleration in the X Direction constant velocity the only equation we have is Delta X equals the X component of velocity times time and again it's that velocity that's constant with no acceleration in the X dimension and plug and chug once again x velocity was our 86.6 meters per second it's not exactly 10 seconds here in this case it's 10.2 seconds and so in this case it will come out just slightly different than our original answer to the first problem we dealt with but in principle it's looking very similar here so 86.6 times 10.2 and we're going to get 883.3 and I'll just round that down to 883. meters cool not so bad when we've already done the one-dimensional X problem an analogous one-dimensional y problem and then we just put them together but again one big key is where do you start well usually the first thing you're going to solve for is how long and again you've got to decide are you getting that from the X Dimension or the Y Dimension and again figure out which dimension is causing the journey to end and again the ball was being pulled back down to the Earth that's due to gravity which operates in the Y Dimension why we use the Y Dimension equations to First solve for time all right the next problem is going to be dealing with free fall sort of but free fall now in two Dimensions so we're gonna have both X and Y components to deal with and stuff like this but it's still going to be analogous to some of the Free Fall problems we did in the last chapter that were one dimensional so question reads a stone is thrown horizontally off a 100.0 meter tall building with an initial velocity of 25 meters per second ignore air resistance two-part question how long does it take for the stone to reach the ground and how far from the building does the stone hit the ground well if we kind of Follow that motion down here so this thing's gonna curve down and finally hit the Earth so and again we've got to deal with the X and Y independently of each other but one thing to note so initially this thing is thrown perfectly horizontally so what that means is that the X component of the Velocity is all of it 25 meters per second and the Y component it doesn't have one so the initial y velocity is zero and I really should say the initial X component is zero I'm sorry is 25 meters per second but I don't have to say the initial it's the entire the velocity of the entire time in the X Dimension there's no acceleration and so the velocity is never going to change it's going to be 25 meters per second horizontally throughout the entire motion again assuming we get to ignore air resistance as we're instructed do but the Y velocity is going to be changing but the initial y velocity is zero notice that it's no different than the case than if we dropped this Stone off this 100 meter tall building its initial y velocity would be zero as well and in fact if you followed the motion all the way down of a stone that's thrown horizontally and one that's dropped you'd find that they would parallel each other and hit the ground at exactly the same time because the Y part of the Velocity is all the same initial y velocity is zero in both cases the final would be the same in both cases and the time to get there would be the same in their both cases the only difference is that one had an x velocity and one didn't but the Y parts of the motion were identical all right so see how this works so again two-part question how long does it take for the stone to reach the ground and how far from the building does the stone hit the ground and again in projectile motion most the time what you're first going to be dealing with is time and again we've got to talk about why does this journey you know why is this Stone's Journey come to an end because it hits the ground why does it hit the ground because of what's going on in the Y Dimension because of gravity pulling it down if you will and so we're going to use that y Dimension set of equations to figure out the time and again in the X Dimension this is all we've got to work with constant velocity no acceleration in the Y Dimension uniform acceleration due to gravity and there's our suite of equations to work with which one do we want to use for time well in this case we have a problem my go-to equation my favorite it's that first one it's not going to work for us here we know the initial y component the velocity of zero we don't know what the final y component of velocity is could we figure it out yeah we could but it's going to be some work to figure it out just so that we could use this equation probably not the best choice in this case then all right what about the second one could we use that one instead uh to figure out the time well the initial y velocity is zero that's nice that first component will go away and so technically we know that the Y displacement negative 100 meters down we know the acceleration due to gravity negative 9.8 meters per second per second we could solve for the time right there as well so that's going to be the way to go and so in this case in that y direction Delta Y is going to equal the initial y component of velocity times time plus one half the acceleration in the y direction t squared all right and once again the initial y component of the Velocity is zero so that first term goes away and we're just left with well we'll start filling some numbers in so and we keep in mind that the displacement is downward and I'm going to write that as negative 100 meters so if you get your signs wrong here and stuff like that you're gonna have some problems we'll see what I mean here in a second so one half acceleration due to gravity is negative 9.8 meters per second squared times t squared if we would have forgot to make this negative but still made acceleration negative we'd have found out we've been trying to take a square root of a negative number when we're all said and done and that's a problem so you got to remember that not only does acceleration Point downward but technically the displacement is in the downward direction for the motion here and therefore that displacement is a negative number as well so signs are kind of important much of the time here all right so if we take 100 and whether you want to divide by one half or multiply by two and then divide by 9.8 or negative 9.8 as the case may be we're going to get 20.4 let's write out that step by step here so t squared equals 20.4 let's put one more on there 4 1. and that's in second squared we'll take the square root of both sides and where do you get 4.52 seconds which in this case is going to round to 4.5 seconds okay first question is answered second question is how far from the building does this Stone hit the ground and so now we want this displacement right here which is the X displacement Delta X well again with no acceleration constant velocity in the X Direction that's the only equation we have to work with we need the X component the velocity times time X component of velocity is constant throughout the entire journey of 25 meters per second we've now figured out that the journey is going to last 4.5 seconds and we can let our calculator go back to work for us and we're going to get 112.5 which in this case if we look back at our input numbers we're going to be limited by the 25 meters per second at two sig figs and around this to two sig figs we'd round it down to 110 meters expressing the proper number of sig figs cool that's the whole problem so total time is 4.5 seconds we use the Y motion to figure that out because it's the Y motion that's causing the journey to end so and then we figured out the X displacement how far from the building it lands as well all right our last projectile motion question here so in this case we've got an arrow fire from a crossbow at a very tall building located 200 meters away with an initial velocity of 98 meters per second at an angle of 30 degrees above the horizontal and the question is at what height does the arrow strike the building so this arrow is going to hit the building well a couple things we don't know we don't know if it's going to hit this building on the way up we don't know if it's going to hit it on the way down and the truth is we don't even actually technically know except for the way the problems worded if the arrow goes over the building and in fact the height of the building's not even given it's just told that it's very tall and in that case it's word in such a way that it's it's tall enough that the arrow is going to hit the building it's not going to go over okay the question is how far up well and the truth is whether we hit it on the way up still or after it hits Max height and comes back down it doesn't matter all we need to figure out is what is the why displacement when this happens and so whether we drew this out that it hits it back on the way back down doesn't matter because in that case it doesn't matter as long as or if we hit it on the way up or something like it doesn't matter on the directory the only thing we really have to worry about is what is the widest placement and whether we're doing it for you know on the way up or coming on back on the way down the widest placement from here would be the same either way we're going to label exactly Delta Y in either case and we'll calculate it exactly the same way as well all right so first thing we want to do is break that velocity up into X and Y components so your X component is going to be 98 meters per second times the cosine of 30 degrees your y component of your velocity is going to be 98 meters per second times the sine of 30 degrees so and again I use sine of 30 because I like the math works the way it works out sine of 30 is one half half of 98 is 49 meters per second for the Y component of our velocity so we'll see that's very convenient number and for the X components 98 times the cosine of 30. it's going to get us 84.9 foreign meters per second okay we're going to start with time and the question we have to ask ourselves is again why does this journey come to an end so well it comes to an end because we hit the building the question is did we hit the building because what happened in the X Dimension or what happened in the Y Dimension well again in the previous problem you know the the Journey of the the stone came to an end uh and the journey of the baseball and the problem before that came to an end because gravity pulled the ball into the ground is gravity pulling the ball into this building per se so well gravity's not pulling anything in a horizontal Dimension so the reason this journey comes to the end is we're hitting the building based on the horizontal motion now not the vertical and that's why I chose this question I wanted you to see an example of that and so it is the X Dimension that's going to deal with why in the world this journey is coming to an end that's where we start to calculate the time well again in the X Dimension there's no acceleration it's constant velocity the only equation we have to work with is this guy and so Delta x equals v x t so we'll solve for that time the X displacement we know that's going to be 200 meters x velocity here is 84. 0.9 meters per second and then times time and we can see well we're traveling 84.9 meters per second horizontally to get a total displacement horizontally of 200 meters it's going to take over two seconds well how long exactly 200 divided by 84.9 is going to be 2.36 seconds okay so now we know how long the journey is the question that is well then what would be the wide displacement after 2.36 seconds has passed okay well this might be one way to go do we know the average y velocity well we know the initial y velocity is 49 meters per second at that point in the journey do we know what the Y velocity is well no we don't so then we're not going to be able to calculate an average y velocity at this point okay move on to the next equation could we use this one do we know the initial y velocity yes it's 49. so do we know the time yes plug it in 2.36 seconds so do we know the acceleration due to gravity in the y direction yes negative 9.8 meter per second squared yeah we could use this one we definitely could so and then we've got this one as well so do we know the initial and final y velocities we don't so we can't use this one the only one we can use is this one at least very easily all right so we got Delta y equals V initial y component times time plus one half acceleration in the y direction times time squared initial y velocity was 49 meters per second time 2.3 seconds 2.36 seconds acceleration due to gravity negative 9.8 meters per second per second or meters per second squared and again times the time of 2.36 seconds and we've got to remember not to forget to square that right there as well and now we can let our calculator do the work and again we're actually not going to know if we're on the way up or if we're on the way down and it doesn't matter we'd solve for Delta y the same way either way all right so 49 times 2.36 plus one half times negative 9.8 really important to get your sign there times 2.36 squared and we're going to get 88.35 meters and if we go back to our inputs here it looks like the 98 meters per second here was giving us two sig figs that's going to be our limiter actually so around this we're going to have to round it here and it's going to round down and so Delta Y is equal to 88 meters that is your final answer now this question doesn't deal with this but my question for you is did we hit this on the way up or did we hit it on the way down as I diagrammed out and how would we figure that out well let's take a look at this for a second so the initial y velocity is 49 meters per second roughly 50 meters per second and it's going to slow down on the way up 10 meters per second every second how long is it going to take us to reach the max height about five seconds in fact exactly five seconds in this problem uh with it being 49 meters per second and gravity being 9.8 meter seconds per second directed downward exactly five seconds and so just to get to the max height would take five seconds well we're way before that and so I didn't diagram this correctly and it didn't matter how we solved it the truth is we're actually going to hit the building before we actually hit the maximum height in the trajectory of that Arrow now if you found this lesson helpful give it a like happy studying