hi everybody physics ninja here today what i want to do is look at free fall problems i'm going to look at three types of problems uh the first problem you could be standing on top of a building and we're simply going to drop an object i want to analyze its position and its velocity as a function of time a second problem make it a little bit harder what i want to do is instead of dropping it i may throw the object up in the air it's going to go up to some maximum height turn around come back down again i want to analyze the motion of the object at any time and the last one is another variation except instead of throwing it up what happens if i throw it down right if i'm at the top of a building and i throw an object down it should take a lot less time to reach the bottom right anyway let's go ahead and analyze these three cases what we're going to do is look at the kinematic equations and how do we apply them to solve free-fall problems again with all my videos please the best way to support what i do is to subscribe to my channel and like the video alright remember leave all your comments and questions down below i'll do the best i can to get back to you as quickly as possible all right let's get started i first want to do a quick refresher on our kinematic equations i'm going to write down three equations and these equations i can write as long as two conditions are meant uh first of all the we're looking at 1d motion and for free fall we will be looking at 1d motion either up or down and we also require that the acceleration is uniform or is a constant value all right and whenever you have both of these conditions i can write simple equations to analyze the motion so i typically use three equations to solve most of these problems the first one here says that the final velocity equals to the initial velocity plus the acceleration times time uh the second one is my displacement is the initial velocity times time plus one-half the acceleration and multiplied by time squared and our last one here is v final squared equals to v initial squared plus two times the displacement delta y multiplied by the acceleration now if i want to write these equations specifically for the free fall problem i only have to do two things okay first of all i simply have to exchange the acceleration here has a constant value for free fall and the value we typically write it like this it's equal to minus little g and little g is defined as the acceleration due to gravity and its value is well known it's 9.8 meters per second squared when you're close to the surface of the earth so that's it okay so we're going to substitute little g into our three equations here and apply these equations to solve all our free fall problems so let me go ahead and just kind of make an arrow here we'll adapt these three equations for free fall so the first one here will simply become v final equals to v initial minus little g times t now you may ask why is there a negative sign there and that's a really really good question when i've written this remember acceleration is a vector quantity and in order to denote the direction for 1d problems you have to pick a direction and by writing minus 9.8 or minus little g i'm assuming that the down direction is a negative direction all right we'll see more of that when we look at problems what else our second equation then would simply become that the displacement in the vertical direction is our initial velocity times time minus one-half little g t squared and then the final one would become v final squared equals v initial squared minus 2 times the displacement times little g all right these are our three equations that we're going to use for all the free fall problems our acceleration is constant and it's pointing down everywhere and it is also has the value of 9.8 meters per second squared let's go now look at some problems all right here's our first case we're going to drop the ball we're going to drop the ball from a height h in this case here i'm gonna set the height equals to seven meters okay kind of a tall building if i drop a ball a very important characteristic of that is that my initial velocity has to be zero meters per second okay what else i've defined the up direction to be positive so if i was going to go everywhere and plot the direction of the acceleration for example what is the acceleration as soon as i drop it i would have to do something like this the acceleration here would be down and the value would be little g and little g is 9.8 now what if i drew it right here the acceleration would also be down and the value would still be little g 9.8 and our last let me do it right before it hits the ground the acceleration would still be 9.8 pointing down what about the velocity well the velocity here is zero what happens actually when it's kind of partially down here again the velocity vector points down right this would be the velocity over here and right before it hits the ground it's also pointing down but i should make that vector a little bit bigger because it's speeding up all right so the velocity starts zero gets bigger and bigger and bigger okay so let's look at a couple standard questions for a type of problem like this how long before it hits the ground right how much time does it take basically to fall seven meters how would you solve that problem all right so let's go ahead and do this so first of all these are the three kinematic equations and let's look at which one i might use why would i not use equation one for how long it hits the ground first of all it has to have time in it right but why not use equation one because i don't know what the final velocity is right here if i knew this then i could use equation one so equation two has time and equation three doesn't have time so i'm probably guessing that equation two is probably the best choice over here initial velocity times time minus one half a little g t squared now some simplifications happen over here first of all this first term has to go to zero because the initial velocity we said was zero whenever you drop something so that becomes straightforward the next thing is what is the displacement for this problem okay remember a displacement is always a final position minus an initial position this is actually how you write any delta value it's always a final value minus an initial value and this here has to be equal to minus one half little g t squared all right so for this we're going to use our displacement well we know it's 7 meters but should i use positive seven or should i use negative seven okay well this kind of depends typically what we do is we set the zero somewhere we're gonna say maybe that the ground is position zero and up here we're actually at position seven meters above that point and it's positive seven right so if i substitute our values over here my final value would be zero my initial value would be minus minus the initial value which is minus seven and that here has to be equal to minus one half little g is 9.8 and t squared you see that this side is the displacement and this side is negative and that's okay because my displacement vector for this problem is actually down it's pointing down and the magnitude is seven meters all right our last thing all we have to do now is do a little bit of algebra so the negative signs cancel out and we put the two at the top we divide by nine point eight and we get t squared equals 2 times 7 over 9.8 and at the end you could take the square root of each side and then you're going to be left with time okay if i do time substituting all our numbers i should get approximately 1.2 seconds okay for the ball to drop a distance of 7 meters all right problem 2 says how fast is it moving right here at this position right before it hits the ground well now it's probably best to use equation one let's try that one right v final that's what i'm trying to solve for equals to the initial velocity minus little g multiplied by time well think about this problem right that initial velocity is zero i know little g is 9.8 and which time would i use in this case here i would use the time that i just solved for because this is how much time it takes before the ball gets to this final position so all you have to do now is simply substitute those numbers all right so you're going to get minus 9.8 multiplied by 1.2 and at the end if you do that carefully oh man i think i did i should get negative 11.7 meters per second okay that is the final speed all right remember this negative sign here is just an indication that the final velocity as a vector is pointing down so we'd expect something to be negative all right and what if we didn't know this and we wanted to use equation 3 instead how do we apply equation 3 to this problem it would look something like this right let's just write it down 2g times delta y again we have that simplification that the initial velocity is zero forget about that term and here now we're just substituting the numbers right so you get minus two little g the magnitude is nine point eight and the displacement over here should be minus seven again negative because the ball's dropping and that's the negative direction for this problem now this is v f squared so notice these two negatives are going to cancel out so let's not worry about that and now if we take our time here we take that the final velocity if you take the square root of something guess what you get two answers plus or minus two times nine point eight multiplied by seven and guess what this equals to plus or minus eleven point seven meters per second at the end if we're looking at an object that is dropping and i've defined down to be uh the negative direction then at the end i have to just keep the negative solution i have to look at what's going on and my velocity should be negative 11.7 meters per second all right this next case is really really important what happens when you throw a ball up all right again we're going to assume here that my height was seven meters same as before all right that's the initial height and i'm going to assume that i'm going to throw this ball up here at a speed of 6 meters per second it's going up initially okay anyway so it's going to go up to some maximum height here turn around then come back down right there is something really really really important about this position here right and this is our top position and it's that the vertical velocity at the top equals to zero it instantaneously comes to rest all right so this here is really really important for these problems so again if you were going to plot acceleration anywhere over here on this trajectory you would have to write it down as minus 9.8 meters per second or 9.8 pointing down and down i've defined as the negative direction again so again let's just write down our coordinate system here so everything going up is defined as the positive direction okay these are the four questions i want to look at for this problem how long does it take to get to the top let's start with that one okay so again to use that one i have to use this fact that at the top the vertical component of the velocity is zero which of these equations has time either the first one or the second one okay now i'm looking for how long i don't know what that height is so i really don't know the displacement to get to the top so using equation two is problematic but problem one sorry equation one should do the job for me let's just first write it down so it has time and that's basically what i'm trying to solve for and now i'm going to use the fact that at the top i know the velocity they don't tell me it's 0 but i know it's 0. all right equals what is the initial velocity for this one this is plus 6 for this problem positive because the initial velocity is pointing up and it's that's the positive direction minus 9.8 that's our acceleration multiplied by time so rearrange this equation right here what you're going to get here is 6 over 9.8 the negative signs will cancel out when you solve that and i should get 0.612 seconds so at 6.612 seconds i'm right here at the top of the trajectory that's it for that one pretty straightforward the next problem says find what this maximum height is all right well for that if i can figure out how high i traveled here in this point 612 seconds i should be able to find what the maximum height is it just depends what you want to do it relative to the ground or relative to the top here but we could find what the displacement is here just this part from the initial position to the top for that here i probably want to use either equation two or equation three they're both going to do the dry job uh for equation two it's simply v i t minus one half little g multiplied by t squared okay let's keep going here my initial velocity we said was six the time now is 0.612 that's when i'm at the top minus one-half 9.8 and again 0.612 and don't forget to square that last one all right so my displacement now that i get for this specific specific problem is 1.84 meters okay so that is the displacement now they're asking me for the maximum height so this is where you got to be a little bit careful okay so this 1.84 is really this distance right here right this is the distance here this is my displacement delta y equals to 1.84 okay so the maximum height again if i'm going to measure it say from the ground then the maximum height should be 7.84 okay so the max height just write down again from the ground okay should be uh the seven meter oh sorry seven meters plus one point eight four which is eight point eight four meters okay uh pretty straightforward the next one is let's find the speed now right before hitting the ground so right before hitting the ground is this position right here right and it's moving pretty fast it's moving downward i'm looking for this value over here so this is for problem three all right so which equation should i use well i can either use equation one or i can use maybe equation three also has the speed right maybe let's do that one right there that one's pretty straightforward how would i use equation 3 to find the speed for this problem well again let's just start by writing down our equation right the equation here is v final squared equals v initial squared minus 2 g and delta y my displacement all right so how do i apply the equation to this problem right here um well one thing i could do is again just substituting in our values and again i want to take the square root on each side in order to really get the expression for v final and whenever i'm i do the square root i have two solutions plus or minus like we indicated before we probably only want to keep the negative solution because we know the object is moving down so the initial velocity is 6 squared and then minus 2 multiplied by 9.8 and now we have to be a little bit careful what is the displacement delta y well this is the displacement delta y i start right here and i finish right here guess what that displacement is right here right this is the delta y that i'm looking for for the entire motion so in this case here i should really have negative seven meters as my total displacement the negative is telling me that it's going down and the magnitude is seven okay if you substitute all the numbers in there you should get a final speed of 13.16 meters per second don't forget the negative sign turn that into a vector to indicate that it is going down all right pretty straightforward problem all right for the second part we want to find the last part rather we want to find the total flight time there are many many ways to do this problem here so i've given myself a little bit more space here but let's think about maybe using equation two here to find the total flight time so let me just write four let's write down this equation and think about how we would apply this equation to find the flight time i'll solve it this is probably the hardest way to solve it we found how much time it takes to go to the top right you could just look at going from the top back down and then add both of those times that's actually much easier to do but let's just do it this way just to practice using this equation so first of all we started over here and we want to end up down here so we know our displacement here is minus seven that we know negative sign because it's going down we know the initial velocity is six and it's positive six we don't know time time will be the total time from going from here all the way to the bottom minus one half a little g we know is 9.8 and multiplied by t squared what i'm going to do now is i'm going to put all the terms on one side of the equation okay so let's go ahead and do that let's bring the 7 on the other side so we're going to zero on the left hand side and here we're going to be left with 7 all right plus 6t and then minus 9.8 over 2 is 4.9 t squared okay what we have over here folks is a quadratic equation okay and in order to solve the quadratic equation you can use the solution for the quadratic equation which says that t is going to be equal to remember it's minus b plus or minus the square root of b squared minus 4ac divided by 2a so think about how you would apply it to this problem or you can use your calculator right some scientific calculators can just give you the answer right away but let me go ahead and just substitute the numbers in here so you get minus 6 for this one what else plus or minus square root b squared would be 6 squared minus 4. the a term is minus 4.9 and the c term is positive 7. so you see these two negative signs are going to cancel out and then at the end you're left with 2a so this is 2 multiplied by minus 4.9 right i haven't done anything um right anything everything i've done here is hopefully pretty straightforward you just got to be careful not to make mistakes all right if you evaluate this term here in the square root that's what i'll do first i'll break it down plus or minus i think i got like 13.16 for this then all of this divided by this becomes 2 times negative 4.9 gives me this now you have to think about this there's two solutions here if i use the positive sign right imagine i first look at the positive sign actually what happens here is you're going to end up getting a positive number in the top divided by a negative number but in this case time is going to be negative all right and we can't have that okay so we're not going to use that sign so instead here what i'm going to do here is i'm just going to use the negative sign and you'll see that all of those negative signs are eventually going to cancel out so if you do that negative sign what you end up getting is that the time the positive time ends up being 1.96 seconds okay and this is really the total flight time in order for my displacement to be negative seven meters right so it's the time to go up and then come back down and end up right here all right not an easy problem to solve but pretty straightforward you just have to be careful with the algebra and getting to the final answer all right the last case i want to look at is what happens when you throw a ball down here's the problem height is seven my initial i'm going to throw down at eight meters per second but remember it's going down right um again once it's going to speed up again once it right before it hits the ground it's probably moving faster than eight meters per second uh first question how much time does it take to drop seven meters if i gave it this initial throw of eight meters per second down so which question or which equation should i use to try to find the time again you can't use the first one because you don't know what the final speed is you can use the second one though right so if i use equation two here it's my displacement is initial velocity times time minus one half gt squared all right we should start getting good at substituting numbers in here my displacement since it's going down i should put negative seven over here now my initial velocity what should i do with that should i just put eight so this is where you have to be careful it's eight but it's also pointing down and down is the negative direction because the coordinate system that we typically use for this is positive is pointing up negative is pointing down all right times t what else minus one half uh little g nine point eight and t squared this is kind of similar to the previous one except there's some different signs in here let me bring the seven over and minus eight t again nine point eight over ends up being 4.9 t squared here's my quadratic equation again which i know the solution t is equal to there's two solutions one of them is positive one of them is negative you just take the positive solution so again it's minus b minus b gives me 8 plus or minus the square root of b squared this is going to be 8 squared minus 4 multiplied by a negative 4.9 multiplied by c which is 7 divided by 2a and this will be 2 multiplied by negative 4.9 all right if i did that square root term correctly here what i should get is 8 plus or minus oops if i evaluated this whole term what i should get i think is like 14.18 all of this over negative 9.8 all right if you think about which sign should you use now right if i use the top sign the positive sign what's going to happen well i'm going to have a positive number in the numerator and a negative number that's not going to work all right so let's forget about that positive sign so what you really should use is the negative sign here all right and if you do that i think that the time you should get at the end is 0.63 seconds that's how much time it takes to drop seven meters if i throw something down all right what about now what's the second question find the velocity just before hitting the ground again there's two approaches i can use to this one right i could use the first equation v final equals to v initial minus little g times t in this case i'm looking for v final so what's my v initial negative 8 minus 9.8 multiplied by 0.63 seconds all right if you did that carefully what you should get is approximately 14.2 meters per second as my final velocity i could have also used equation 3 right what if i used equation 3 instead right if you use the equation 3 that also has the final velocity again if you take the square root on each side you're ending up just substituting numbers here we'll have to take the negative solution as we indicated before little g multiplied by delta y okay so let's just go right away and cancel out that positive solution and what i'm left with here is my final velocity equals two this will be eight squared minus 2 times 9.8 now my displacement again is going to be negative 7 meters substituting the numbers in there again there may be a little bit of rounding error here but should give me approximately 14.2 if you take care of all the rounding these answers should be exactly identical to each other okay all right folks that's it for my free fall video tutorial hopefully you found these problems useful uh thanks for watching