Question

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

(a) What is the height of the ball to the nearest tenth of a meter exactly 1 seconds after he drops the ball?
(b) How many seconds, after the ball is released, will it hit the ground?

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 - (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 - (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 - (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 - (1)/(2)gt^2`

And solving for t we find

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