1. The problem statement, all variables and given/known data Julie Throws a ball to her friend sarah. The ball leaves julies hand a distance 1.5m above the ground with an initial speed of 23m/s at an angle of 26° wrt the horizontal. Sarah catches the ball 1.5m above the ground. (a) what is the horizontal component of the balls velocity when it leaves julie's hand (b) what is the vertical component of the balls velocity when it leaves julie's hand (c) what is the maximum height the ball goes above the ground? (d) what is the distance between the two girls? (e) After catching the ball sarah throws it back to julie. The ball leaves sarah's hand a distance of 1.5m above the ground, and it is moving with a speed of 25m/s when it reaches a maximum height of 9m above the ground. What is the speed of the ball when it reaches sarahs hand? (f) how high above the ground will the ball be when it gets to julie? 2. Relevant equations xf = xi + v0t +.5at^2 vf^2 = vi^2 + 2a(x - x0) v = v0 + at 3. The attempt at a solution I got to the last part but I want to check my work over the previous parts, there are no answers in this book. for (a) and (b) I have vx = 23cos(26°) and vy = 23sin(26°) for (c) i want to know the final height over some interval that is not the entire distance, I had to calculate the total time that the ball was in the air which I got using yf = yi +v0yt +.5at2 with yf = yi = 1.5m I got 0 = 10.08253638(T)m/s - 4.905m/s2t2m/s2 and the total time as 2.055562922 seconds I usually can divide a lot of these total distances and total times by 2 because of the symmetry of the parabolic path the ball travels. So I went ahead and found t/2 = 1.027781486 so yf(t/2) should give me the maximum height of the ball yf = 1.5m + 10.08253638(1.1.02778148656) + .5(-9.81m/s^2)*(1.027781486)^2 yf = 7.43m okay the next question asks me to find the total distance xf = xi + v0x + .5at2 where xi = 0m + v0x = 23cos(26°) t = 2.05...seconds and for "a" I still used the adtg as 9.81m/s^2 so xf = 63.22m about 180 feet, this one is really long(probably because ax = 0 there is constant velocity in the x, maybe the real distance is 41.34m(got this by just plugging in ax = 0) okay (e) calculate the initial speed of the ball when it leaves sarah's hand okay I need vxi and vyi to get at this initial speed. I tried using vfx = vix + at but vfx = vix so I actually can't use the work I wrote down, looking at it again I must concede I did that part incorrectly, which leaves me at (e) i am not sure how to do this part. maybe I could start by finding vx at the total distance or something?