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?

a) 70.1 m

The ball is moving by uniformly accelerated motion, with constant acceleration (acceleration due to gravity) towards the ground. The height of the ball at time t is given by the equation

where 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:

b) 3.9 s

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

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

And solving for t we find

x coordinate is none

step-by-step explanation: