A rookie quarterback throws a football with an initial upward velocity component of 12.0 m/s and a horizontal velocity component of 20.0 m/s. Ignore air resistance. (a) How much time is required for the football to reach the highest point of the trajectory? (b) How high is this point? (c) How much time (after it is thrown) is required for the football to return to its original level? How does this compare with the time calculated in part (a)? (d) How far has the football traveled horizontally during this time? (e) Draw x-t, y-t, vx-t, and vy-t graphs for the motion.

Solution 12E Consider the ball is projected upward and take this direction as positive y axis The ball is travelling upward so the components of acceleration are a x0 a y-g( - ve sign represents that ball is moving upward means opposite to the direction of acceleration gravity) v y0(at the highest point) a)To find the time, we have equation v =12 m/s 0y 2 g= 9.8 m/s v yv +0yt y 0=v -0y v0y t= g 12 = 9.8 =1.224 s b)Different constant acceleration gives different expressions but the numerical result isa same 2 v0y 2gt = 2 v0y t = 2g =7.35 m c) The time is twice that found in ( a) =2×1.224 =2.448 s d)a x0 v = 20 m/s 0x t=2.448 s 2 x-x 0v t+0x 2 axt x-x =v t 0 0x =20×2.448 =48.96 m49 m e) Graph of x versus t Graph of y-t, Graph of v xersus t Graph of v veysus t