Question

A tennis ball is it at ground level. The ball reaches its maximum height above the ground level 1 sec. after being hit. Then 0.9 seconds after reaching its maximum height, the ball barely clears the net that is 3 m from where the tennis ball was hit. Assume the ground is level. How high is the net?  How far beyond the net does the tennis ball strike the ground?

Answer 1

Let magnitude of initial velocity of ball = v and it is hit at an angle with respect to horizontal

Horizontal component of initial velocity of ball = v cos

Vertical component of initial velocity of ball = vsin

It reaches maximum height in time t = 1 s

At maximum height vertical velocity v' = 0

Using

v' = vsin + at

0 = vsin -9.81(1)

vsin = 9.81 m/s

Also

v'^2 - v^2 = 2ay

0^2 - (vsin)^2 = 2(-9.81)y

- (9.81)^2 = 2(-9.81)y

y = 4.905 m

Now 0.9 s after reaching maximum height ball clears net

We can find vertical displacement y' of ball in this time

y' = v't + 0.5at^2

y' = 0 + 0.5(-9.81)(0.9)^2 = -3.97 m

Thus, height of net above ground = Maximum height reached by ball - vertical distance covered by ball in 0.9 s after reaching maximum height

height of net = 4.905 - 3.97 = 0.935 m

Now, tennis ball hits ground x distance far beyond net

x distance is covered by ball in 0.1s ( 1s - 0.9s = 0.1s) as ball takes the same time to fall down as it takes to reach maximum height

x = horizontal velocity x 0.1 s = vcos x 0.1

vcos can be calculated using the information that ball covers 3 m distance in 1.9s ( 1s to reach maximum height + 0.9 s further to reach net)

3 = v cos x 1.9

vcos = 1.58 m/s

Therefore, x = 1.58 x 0.1 = 0.158m