Answer to Question #137550 in Physics for Raymond J Carroll

Question

A steel ball rolls with a constant velocity across a table top 0.808 m high. It rolls off and hits the ground +0.264 m horizontally from the edge of the table. How fast was the ball rolling?

Answer 1

Let's first write the equations of motion of the steel ball in horizontal and vertical directions:

$$x = vt,$$ $$y = \dfrac{1}{2}gt^2,$$

here, $x = 0.264 \ m$ is the horizontal displacement of the ball, $v$ is the ball's velocity across the table, $t$ is time that the ball takes to reach the ground, $y = 0.808 \ m$ is the vertical displacement of the ball (or the height of the table) and $g = 9.8 \ \dfrac{m}{s^2}$ is the acceleration due to gravity.

Let's find the time that the ball takes to reach the ground from the second equation:

$$t = \sqrt{\dfrac{2y}{g}}.$$

Then, we can substitute $t$ into the first equation and find $v$:

$$x = v\sqrt{\dfrac{2y}{g}},$$ $$v = \dfrac{x}{\sqrt{\dfrac{2y}{g}}} = \dfrac{0.264 \ m}{\sqrt{\dfrac{2 \cdot 0.808 \ m}{9.8 \ \dfrac{m}{s^2}}}} = 0.65 \ \dfrac{m}{s}.$$

Answer:

$v = 0.65 \ \dfrac{m}{s}.$