Question

David drops a soccer ball off a building. The building is 75 meters tall.

Answer 1

a) 70.1 m

The ball is moving by uniformly accelerated motion, with constant acceleration $g=9.8 m/s^2$ (acceleration due to gravity) towards the ground. The height of the ball at time t is given by the equation

$h(t)=h_0 - \frac{1}{2}gt^2$

where $h_0 = 75 m$ is the height of the ball at time t=0. Substituting t=1 s, we can find the height of the ball 1 seconds after it has been dropped:

$h(1 s)=75 m - \frac{1}{2}(9.8 m/s^2)(1 s)^2=70.1 m$

b) 3.9 s

We can still use the same equation we used in the previous part of the problem:

$h(t)=h_0 - \frac{1}{2}gt^2$

This time, we want to find the time t at which the ball hits the ground, which means the time t at which h(t)=0. So we have

$0=h_0 - \frac{1}{2}gt^2$

And solving for t we find

$t=\sqrt{\frac{2h_0}{g}}=\sqrt{\frac{2(75 m)}{9.8 m/s^2}}=3.9 s$