Question

David drops a ball from the top of a building with a height of 64 feet.

Answer 1

Answer:

Step-by-step explanation:

The position formula for this is actually

$s(t)=-16t^2+64$

It says, in not so may words, that the object was dropped under the influence of gravity from a height of 64 feet. When it hits, the ground, it will not be above ground at all anymore...it will be ON the ground and have a height of 0. Therefore, if we set the position function equal to 0 and solve for t, we are finding out how long it will take for the object to hit the ground.

$0=-16t^2+64$ and

$16t^2=64$ and

$t^2=4$ and

t = 2

It takes 2 seconds for the object to hit the ground from 64 feet up.