Asked by anika

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?
Answered by Steve
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
Answered by dem
what is acceleration ?
Answered by dem
what is distance?
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