Question

A ball is tossed so that it bounces off the ground, rises to a height of 2.50 m, and then hits the ground again 0.90 m away from the first bounce.
1.How long is the ball in the air between the two bounces?2.What is the ball's velocity in the x-direction?3.What is the ball's speed just before the second bounce?4.What is the angle of the velocity vector with respect to the ground right after the first bounce

Answer 1

To solve this problem we will apply the linear motion kinematic equations. With the equation describing the position with respect to acceleration (gravity), initial velocity and time, we will find time.

Our values are,

$h = 2.5m$

$x = 0.9m$

1) We know that, the time, the ball in the air depends on the maximum height.

Therefore the height is calculated as

$h = v_{yi}t+\frac{1}{2} gt^2$

There is not initial velocity then replacing,

$2.5 = 0t + \frac{1}{2}9.8t^2$

$t = \sqrt{\frac{2(2.5)}{9.8}}$

$t = 0.714s$

By symmetry, the time taken to reach ground is equal to the time taken to reach the maximum height, then

$T = t+t$

$T = 2*0.714s$

$T = 1.426s$

2 ) In the time t=1.428s, the ball moves horizontally a distance x = 0.9

Then the horizontal velocity,

$v_x = \frac{x}{t} = \frac{0.9m}{1.428s}$

$v_x = 0.6302m/s$

3 ) The ball's speed just before the second bounce is found as

$v_y = v_{yi} +gt_{down}$

Replacing we have

$v_y = 0+(9.8)(0.714)$

$v_y = 6.9972m/s$

4 ) The angle of the velocity vector with respect to the ground right after the first bounce is

$\theta = tan^{-1} (\frac{v_y}{v_x})$

$\theta = tan^{-1} (\frac{6.9972}{0.6302})$

$\theta = 84.8\°$