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

Answer:

5.1 m/s

Explanation:

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

:

$s=ut+\frac{1}{2}at^2$

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:

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(1.1)}{9.81}}=0.474 s$

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

$v_x = \frac{d}{t}=\frac{2.44}{0.474}=5.1 m/s$

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.