Question

What is the ejection speed of a ball if it was shot horizontally 1.1 meter above the floor, and touched it 2.44 meters further from the vertical projection of the ejection point? Use 9.810 m/s^2 for the acceleration of gravity.

Answer 1

5.1 m/s

Step-by-step explanation:

First we consider the vertical motion of the ball, which is a uniform accelerated motion (free fall). So can use the suvat equation

:

`s=ut+(1)/(2)at^2`

where

, choosing downward as positive direction:

s = 1.1 m is the vertical displacement

u = 0 is the initial vertical velocity of the ball

t is the time

`a=g=9.81 m/s^2` is the acceleration of gravity

Solving for t, we find the time at which the ball hits the ground:

`t=\sqrt{(2s)/(g)}=\sqrt{(2(1.1))/(9.81)}=0.474 s`

Now we consider the horizontal motion. This is a uniform motion with constant horizontal velocity. Therefore, the horizontal distance travelled by the ball is given by

`d=v_x t`

where

d = 2.44 m is the horizontal distance travelled (we are told that the ball hits the ground 2.44 m further from the vertical projection of the ejection point)

t = 0.474 s is the time of flight

Solving for
`v_x` we find

`v_x = (d)/(t)=(2.44)/(0.474)=5.1 m/s`

The horizontal velocity is constant during the whole motion (since there are no forces acting along this direction), so this is the ejection speed of the ball.