A ball is tossed so that it bounces off the ground, rises to a height of 0.90 m, and then hits the ground again 0.50 m away from the first bounce.

How long is the ball in the air between the two bounces?


What is the ball's velocity in the x-direction?


What is the ball's speed just before the second bounce?


What is the angle of the velocity vector with respect to the ground right after the first bounce?

3 answers

s = 1/2 at^2
time to rise .9m:
.9 = 1/2 (9.8) t^2
t = .428 sec

.5m/.428s = 1.17m/s

v = at = 9.8*.428 = 4.19m/s

h = -14.4x^2 + 7.2x
h' = -28.8x + 7.2
h'(0) = 7.2
arctan(7.2) = 82 degrees
what is acceleration ?
what is distance?