Answer to Question #91699 in Classical Mechanics for Louie

Question

A ball is left to drop, from rest, vertically from the top of a building from a position 48 m above ground.
i. What is the distance travelled by the ball after it has been falling for 3 seconds?
ii. What is the downwards velocity of the ball as it strikes the ground?

Answer 1

The ball is moving starting from rest with constant acceleration $g = 9.8\, \text{m/s}^2$. The distance $s$ and velocity $v$ after time $t$ are given, respectively, by the relations

$s = \frac12 g t^2 \, , \qquad v = g t \, .$

Expressing the time $t = \sqrt{2 s/g}$ from the first relation, and substituting it into the second one, we obtain the velocity as a function of the distance:

$v = \sqrt{2 g s} \, .$

Substituting $t = 3\, \text{s}$ into the first relation, and $s = h = 48\, \text{m}$ into the last one, we obtain the answers to the problem:

i. $s = 44.1\, \text{m}$

ii. $v = 30.7\, \text{m/s}$