Question

A ball dropped from the top of a building has a height of s =576- 16t2 meters after t seconds. How long does it take the ball to reach the ground? What is the ball's velocity at the moment of impact?

Answer 1

Answer:

It will take 6 seconds for the ball to reach the ground.

Velocity: $-192\frac{\text{m}}{\text{s}}$.

Step-by-step explanation:

We have been given that a ball dropped from the top of a building has a height of $s =576-16t^2$ meters after t seconds.

The ball will hit the ground, when height will be 0 meters, so we will equate height with 0 as:

$576-16t^2=0$

Let us solve for t.

$576-576-16t^2=-576$

$-16t^2=-576$

$\frac{-16t^2}{-16}=\frac{-576}{-16}$

$t^2=36$

Taking positive square root of both sides, we will get:

$\sqrt{t^2}=\sqrt{36}$

$t=6$

Therefore, it will take 6 seconds for the ball to reach the ground.

To find the ball's velocity at $t=6$, we will take the derivative of position function and evaluate derivative at $t=6$.

$s=576-16t^2$

$s'=\frac{d}{dt}(576)-\frac{d}{dt}(16t^2)$

$s'=0-32t$

$s'(6)=-32(6)$

$s'(6)=-192$

Therefore, the ball's velocity at the moment of impact would be $-192\frac{\text{m}}{\text{s}}$.