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

0.79 m/s

Step-by-step 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+(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{(2s)/(a)}=\sqrt{(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 = (d)/(t)=(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.