Question

A ball rolls over the edge of a platform with only a horizontal velocity. The height of the platform is 1.6 m and the horizontal range of the ball from the base of the platform is 20 m. What is the horizontal velocity of the ball just before it touches the ground? Neglect air resistance. Use g = -9.8 m/s². a. 4.9 m/s b. 9.8 m/s c. 35 m/s d. 20 m/s e. 70 m/s

Answer 1

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

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

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

$a=g=9.8 m/s^2$ is the acceleration of gravity

Solving for t, we find the time the ball takes to reach the ground:

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(1.6)}{9.8}}=0.57 s$

Now we can consider the horizontal motion, which is a uniform motion with constant speed. The horizontal distance travelled is given by

$d=v_x t$

where

d = 20 m is the horizontal distance travelled by the ball

t = 0.57 s is the time of flight

Solving for vx,

$v_x = \frac{d}{t}=\frac{20}{0.57}=35 m/s$

And since the horizontal velocity is constant, this is also the horizonal velocity of the ball just before hitting the ground.