# A ball rolling at a constant speed falls off of a table and lands 98 cm from the bottom of the table, 0.49s after rolling off of the edge. How fast was the ball rolling on the table? How tall is the table? How far from the table is the ball after 0.061s?

Sep 16, 2015

$\left(a\right) \text{ "2"m/s}$

$\left(b\right) \text{ "1.17"m}$

$\left(c\right) \text{ "0.122"m}$

#### Explanation:

$\left(a\right) .$

The horizontal component of the velocity remains constant so:

$v = \frac{\Delta s}{\Delta t} = \frac{98}{0.49} = 200 \text{cm/s"=2"m/s}$

This must be the velocity over the table since this is constant also.

$\left(b\right) .$

For the vertical component the ball is falling from rest under constant acceleration $g$.

$s = \frac{1}{2} \text{g"t^2=1/2 xx9.8xx0.49^2=1.17"m}$

$\left(c\right) .$ The horizontal component of velocity is constant so:

$s = v \times t = 200 \times 0.061 = 12.2 \text{cm" =0.122"m}$.