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
:
[tex]s=ut+\frac{1}{2}at^2[/tex]
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
[tex]a=g=9.81 m/s^2[/tex] is the acceleration of gravity
Solving for t, we find the time at which the ball hits the ground:
[tex]t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(1.1)}{9.81}}=0.474 s[/tex]
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
[tex]d=v_x t[/tex]
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 [tex]v_x[/tex] we find
[tex]v_x = \frac{d}{t}=\frac{2.44}{0.474}=5.1 m/s[/tex]
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.
