A 1.5 m tall woman throws a tennis ball horizontally. The ball lands 20 m away. What was the initial horizontal velocity of the ball?

1 Answer
Aug 12, 2017

In projectile motion, we can treat the horizontal and vertical directions separately. Doing so in this case leads to the answer #u_x=36.4# #ms^-1#.

Explanation:

Let's summarise what we know, and what we want to know:

Initial horizontal velocity: #u_x=?# #ms^-1#
Final horizontal velocity, #v_x#, will be the same (ignoring air resistance)
Initial vertical velocity: #u_y=0# #ms^-1#
Final vertical velocity, #v_y#, we don't particularly care about for this question.
Time: #t=?# #s#
Horizontal distance: #s_x=20# #m#
Vertical distance: #s_y=1.5# #m#
Vertical acceleration: #a_y=9.8# #ms^-2# (due to gravity)

First step is to calculate the time taken for an object to fall #1.5# #m# due to gravity:

#s_y=u_yt+1/2a_yt^2#

Since #u_y=0#, that term disappears. We can rearrange the equation to make #t# the subject:

#t=sqrt((2s_y)/a_y)=sqrt((2xx1.5)/9.8)==sqrt((3)/9.8)=0.55# #s#

Second step is to realise that the horizontal velocity is unchanged, and that the ball covered #20# #m# in #0.55# #s#.

#v_x=u_x=s_x/t=20/0.55=36.4# #ms^-1#