Question

A ball is thrown in the air so that its height h after t seconds is given by the function h(t) = -16t^2 + 144t. Find the number of seconds the ball is in the air when it reaches a height of 128 feet. After how many seconds will the ball hit the ground?

Answer 1

a. 1 second

b. 9 seconds

Step-by-step explanation:

Part A.

To find the time when the ball reaches the height of 128 ft, we put h(t) = 128 into our equation and get

`128=-16t^2+144t`

We rearrange the above equation to write it in a more familiar form by subtracting 128 from both sides

`-16t^2+144t-128=0`

The above equation is quadratic, and therefore, we can solve for t using the quadratic formula.

The quadratic formula gives

`t=\frac{-144\pm\sqrt[]{114^2-4(-16)(-128)}}{2(-16)}`

which upon simplification gives us the following solutions.

`\begin{gathered} t=1 \\ t=8 \end{gathered}`

Hence, 1 and 8 seconds after the throw, the ball is at a height of 128 ft.

Since we want the earliest time, from the two values of t we choose t = 1 as our relevant answer.

Part B.

When the ball hits the ground, its height above the ground is zero. Therefore, we put h= 0 into our formula and get

`0=-16t^2+144t`

To solve for t, we first add 16t^2 to both sides and get

`16t^2=144t`

dividing both sides by 16t gives

`(16t^2)/(16t)=(144t)/(16t)`
`\boxed{t=9.^{}}`

Hence, the ball hits the ground 9 seconds after the launch.