Answer:
c. 35 m/s
Explanation:
We start by analzying the vertical motion of the ball, which is a free fall motion, so a uniform accelerated motion. Therefore we can use the suvat equation
[tex]s=ut+\frac{1}{2}at^2[/tex]
where
s = 1.6 m is the vertical displacement (we chose downward as positive direction)
u = 0 is the initial vertical velocity of the ball
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 the ball takes to reach the ground:
[tex]t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(1.6)}{9.8}}=0.57 s[/tex]
Now we can consider the horizontal motion, which is a uniform motion with constant speed. The horizontal distance travelled is given by
[tex]d=v_x t[/tex]
where
d = 20 m is the horizontal distance travelled by the ball
t = 0.57 s is the time of flight
Solving for vx,
[tex]v_x = \frac{d}{t}=\frac{20}{0.57}=35 m/s[/tex]
And since the horizontal velocity is constant, this is also the horizonal velocity of the ball just before hitting the ground.
