Question

A steel ball rolls with constant velocity on a tabletop 1.95 m high. It rolls off and hits the ground 0.5 m away from the edge of the table. How fast was the ball rolling?

Answer 1

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

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

where

s = 1.95 m is the displacement

u = 0 is the initial vertical velocity

t is the time

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

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

$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(1.95)}{9.8}}=0.63 s$

Now we can analyze the horizontal motion: this is a uniform motion with constant speed, so the horizontal distance covered by the ball is

$d=v_x t$

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:

$v_x = \frac{d}{t}=\frac{0.5}{0.63}=0.79 m/s$

And this velocity is constant during the motion, so the ball was moving at 0.79 m/s when it rolls off the table.