Answer:
0.79 m/s
Explanation:
First of all, we analyze the vertical motion of the ball. It is a free fall motion, so its vertical displacement is given by
[tex]s=ut+\frac{1}{2}at^2[/tex]
where
s = 1.95 m is the displacement
u = 0 is the initial vertical velocity
t is the time
[tex]a=g=9.8 m/s^2[/tex] is the acceleration of gravity
Solving for t, we find the time it takes for the ball to reach the ground:
[tex]t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(1.95)}{9.8}}=0.63 s[/tex]
Now we can analyze the horizontal motion: this is a uniform motion with constant speed, so the horizontal distance covered by the ball is
[tex]d=v_x t[/tex]
where
d = 0.5 m is the horizontal distance covered
t = 0.63 s is the time
Solving for vx, we find the horizontal velocity of the ball:
[tex]v_x = \frac{d}{t}=\frac{0.5}{0.63}=0.79 m/s[/tex]
And this velocity is constant during the motion, so the ball was moving at 0.79 m/s when it rolls off the table.
