Question

A ball is thrown into the air. The path of the ball is represented by the equation h(t)= -t^2+8t. How long is the ball in the air?

Answer 1

Answer:

The ball is 8 seconds in the air.

Step-by-step explanation:

Equations

The height of a ball thrown in the air is given by the equation:

$h(t)=-t^2+8t$

We assume the time is expressed in seconds.

The ball will be in the air while its height is positive. When its height is zero, it means the ball could be at launch time or returning to the ground level.

Setting h(t)=0:

$-t^2+8t=0$

Factoring:

$t(-t+8)=0$

There are two possible solutions:

t=0, t=8 sec

The first solution corresponds to the moment the ball is being thrown, the second solution corresponds to the moment the ball is returning to the ground, thus:

The ball is 8 seconds in the air.