[Music] all right so that's basically General motion um next is basically a special case of General motion which is projectile motion so projectile motion is essentially anything that is acting under uh Earth gravity or acceleration due to gravity so there's some key um I point that's important in understanding projectile motion um so if we just consider this drawing right here where we have a cannon that shooting a cannonball horizontally let's just assume for this that gravity is acting downwards in this direction so if we just showed the green ball going to point B would be the actual path that the ball follows and then each of those lines just represents a snapshot in time and the intervals in time are equal so let's draw let's just write delta T delta T delta T T delta T so each of these times are equal so in the horizontal Direction which is perpendicular to the direction acceleration due to gravity so velocity is constant so in this horizontal Direction our velocity is constant it's not changing there's no acceleration term Z so it doesn't matter how fall how far something uh drops I guess the height of the projectile falling um the component of the Velocity in the horizontal direction is always going to be constant and then because we have a constant acceleration G in the vertical direction for the same time increments the distance that the project moves over that time increment is going to be continually increasing and increasing and increasing and that increase is parabolic so the vertical velocity changes due to gravity so those are the main points about projectile motion velocity in the horizontal direction is constant and velocity in the vertical Direction changes due to gravity so if we were to draw just a particle along the path which is undergoing projectile motion just like we did before we can do everything the same as our previous problems except now we just have a constant acceleration term um and we know its component so we have our position vector and that's pointing to our particle which has velocity components BX and b y its velocity is tangent to uh the path it doesn't really look like it's tangent to the path at that point um and we can write our equations in motion so since the X component of our acceleration is equal to zero using our kinematic equations from lecture one we can write the position uh because we have a constant acceleration case the position is equal to some initial position a reference position x0 plus the initial x velocity times time and we only care about the initial X veloc because the Velocity in the X direction is constant it's not changing so we can say the 0x is equal to VX write that up here isal to b0x so the X direction is quite easy we have this one equation uh we can use that if we look at the vertical Direction the Y direction we now have our component a y of our acceleration is equal to minus G and it is constant we said before G is equal to 9.81 m/s squared so we again use those equations of motion uh for constant acceleration we can rewrite the three equations of motion for a projectile as one our y component of velocity is equal to our initial y component of velocity minus acceleration due to gravity times time our next equation our y position is equal to our initial y position plus b0 y * t - 12 acceleration to gravity * t^2 and then our last equation equation 3 V y^ 2 is equal to v0 y^ 2us 2 time acceleration due to gravity times our change in position Yus y0 so let's do a quick example with some projectile motion we have Lon Messi here he kicks a ball at an angle of 30 degrees with respect to the ground and the speed of the ball as it leaves Lionel's foot is 10 meters per second the question says determine how far the ball travels before hitting the ground we have our four equations of motion that we can use one in the X Direction and three in the y direction basic we have to determine the unknown here which is what will be that X position how far will the ball travel before hitting the ground so how far away is point C from point A so basically we're asked to find find XC when ball hits the ground so how can we do this what do we need to know in order to solve this so we have v0x v0 y uh we know the acceleration so it's just gravity um so what is our X component of velocity or initial velocity 10 uh cos 30 and then we know our y component is 10 sin3 so using these if we look back up to our equations um which of these equations do we think might be useful so if we look at equation one we don't really need to find this component of velocity we don't really care um what that is so let's not use equation one if we look at equation three we have that v y again so maybe not equation three so let's try equation two we have a position Y which is convenient just by the geometry it looks like it's going to land at the exact same position and Y as when it was kicked so we know if we use equation two this y over here is going to be equal to Y zero and then we have no other unknowns except for T and then if we have that to be solved for T and we can use that in our equation for our X position we know our b0x we know our x0 say zero we substitute T in there then we can solve for that unknown X so we have two equations we need to solve T and then we can solve for x so let's use that equation for y with the velocity the initial velocity in the y direction is 10 * sin 30 so if we write that we know and write our equation for y y is equal to y z y z plus v0 YT - 12 GT ^2 so Y and Y Z is just going to be equal to uh we'll say y z is equal to zero this is also going to be equal to zero because we're just assuming that that just right this is zero so we're left with P0 y * time is equal to GT ^2 and if we do this we can cancel out one of the t's and we're left with v0 Y so 10 sin 30 is equal to 12 * 9.81 T and we should get T is equal to 1.01 seconds so we've got the amount of time now all we need to do is figure out after 1.01 n seconds how far has that ball moved to the X we can write our equation x is equal to x0 plus b0x time say x0 is zero our b0x is 10 cosine 30 and our time 1.09 should get our X position is equal to 8.83 right that b Neer 8.83 meters so if it's kicked to the velocity of 10 m/s an angle of 30° from the ground it's only going to go 8.83 M before it hits the ground again so let's look at an example which a little bit more complicated so in this case we have a Canon shown it gives us a representation of the inclination of that uh initial velocity and it tells us the initial speed is 150 m/ second so it's asking determine the range of the cannon shown if it has a muzzle velocity and inclination angle of 150 m/s and uh we call the gradient is 3 over4 the rise or run of the inclination is 3 over four and then it also asked how long before the Cannonball lands at B so how should we move ahead with this I always think it's easiest to do the exact same process we did first problem let's just start with some of the things we know so we know v0x is equal to we use similar triangles the X component would just be 4 over 5 time 150 that is meters per second you know v0 y is going to be equal to 3 over 5 150 meters per second and we know that if y z is up here and we can write y b y down here is equal to minus 150 so let's write our yal y0 plus v0 y * t - 12 GT 2 minus 150 is equal to 0 plus our y component uh 3 over 5 * 150 what's 3 over 5 * 150 let's simplify that hopefully you get 90 times t - 12 * 9.81 * t^2 we only have one unknown and if we solve for T so if we solve this equation right here we have a quadratic equation T and t^2 we should get two values for t one of them this is the one we're looking for when the ball lands at point B the other one is going to be some negative amount of time so if we to go backwards through time or move this path backwards it would be the time corresponding to when this Parabola would cross that ground level AIS but we don't care about that negative time we only want the posi value so if we solve this equation and we only look at the positive value we should get T is equal to 1989 seconds so if we use the same process as before we now have time we have our initial X component of velocity so we can write X is equal to x0 which is just going to be zero plus v0x * time so we have four FS of 150 time time which is 1989 and we get X which is our range and that range will be equal to 2,300 and 87 met so if a cannon launched at 150 meters per second at that angle and it's 150 meters above the ground the range of that Cannon or the distance it'll travel before hitting the ground is just about 2400 meters good so in this problem we have Phil is chipping onto the green and he hits his ball with the path that's shown so he's hitting it with a speed of 80 feet per second at an angle of 45 degrees with respect to the ground and the ground is at an angle at 10 degrees with respect to the horizontal how far does Phil's ball get for the first bounce basically what is that distance D when the ball lands at point B so this time it's convenient if we since everything's tilted let let's be I guess specific about uh what coordinate frame we're going to choose because it's projectile motion and gravity is always acting vertically it doesn't really make sense have um our coordinate system on an incline with the ground because if it's on an incline with the ground then we have acceleration in both X and Y directions so let's keep the same equations we had before four let's draw our coordinate reference frame Y is here and X is horizontal so we're going to keep Y is vertical X is horizontal so everything now is going to be with respect to this Frame so we know that b0x is going to be [Music] 80 cosine of the angle between the ball being hit and the ground so it's 45 + 10 so that's 55 degrees v0 Y is going to be 80 * the S of 55 degrees so the other thing we needed to know to solve these problems for is the value um what is that value of y how can we figure this out well y would just be the vertical rise of the ground at the point that it lands at Point D so we're going to find y as a function if we extend this line over here I guess this line This distance y here is going to be D times the sign of the angle of inclination to 10 degre so if we use our equation then we want to solve the time time of flight and then once we know the time of flight we can figure out okay what is the distance traveled in the X direction to reach that and then at the last point we have to not just figure out X we need to figure out the distance D so it's going to be square root of the sum of y^2 + x^2 so let's write our equation Y is equal to y0 plus v0 y t- 12 G t^2 so we'll set this up y now is D 91 we'll say Y is0 is zero our v0 in the y direction is 80 s 55 time T minus 12 acceleration due to gravity in per second squar is 32.2 t^2 so this is one equation you have two unknown still D and T so let's look at our Motion in the X Direction now if we write down X is equal to x0 plus v0x T will end up with X is equal to well x0 will just say is equal to zero plus 80 cosine 5 T and we can write the value of x if the distance along that plane is D that means the distance X we can write as D cosine so we have our second equation so we have two equations two unknowns we can substitute equation two into equation one and then we can solve for that distance and if if we do that we would get D is equal to0 or 166 ft you know obviously the 166 feet is the meaningful answer the one that we care about [Music]