Question

A rocket is launched straight up from the ground with an initial velocity of 192 feet per second. The equation for the height of the rocket at time t is given by:

h=−16t2+192t

a. Find the time when the rocket reaches 560 feet.
b. Find the time when the rocket completes its trajectory and hits the ground.

Answer 1

The equation for the height of the rocket at time t given

`h= -16t^2+192t`

We have to find the time t, when the rocket reaches 560 feet.

That means we have to find t when h = 560 ft. we will place 560 in the place of h to find t now.

`h= -16t^2+192t`

`560 = -16t^2+192t`

In the right side, we can check -16 is the common factor. So we will take out -16 from the rigbht side.

`560 = -16(t^2 - 12t)`

To get rid of -16 from the right side and move it to left side, we will divide both sides by -16.

`560/-16 = -16(t^2-12t)/-16`

`-35 = t^2 -12t`

Now we will move -35 to the righ side by adding 35 to both sides.

`-35+35 = t^2-12t+35`

`0 = t^2 -12t+35`

`t^2-12t+35 = 0`

We will factorize thee left side to find the values of t now. We need to find a pair of factors of 35 that by adding them we will get -12.

The pair of factors of 35 are -5 and -7 and by adding -5-7 we will get -12.

`t^2-12t+35 =0`

`(t-5)(t-7) =0`

So by using zero product property we will get

`t-5 =0`

`t-5+5 = 0+5`

`t=5`

Also
`t-7 =0`

`t-7+7 = 0+7`

`t=7`

So we have got the rocket reaches at 560ft when t = 5 seconds and also when t = 7 seconds.

Now part b.

When the rocket completes its trajectory and hits the ground then the height or h = 0. So we will place h = 0 there in the equation.

`h= -16t^2+192t`

`0= -16t^2 + 192 t`

`0 = -16(t^2-12t)`

`-16(t^2-12t) = 0`

We will move -16 to the other side by dividing it to both sides.

`-16(t^2-12t)/-16 = 0/-16`

`t^2-12t = 0`

We will take out the common factor t from the left side. By taking out t we will get,

`t(t-12) = 0`

We will use zero product property now. By using that we will get,

`t = 0`

ans also
`t-12 = 0`

`t-12+12 = 0+12`

`t = 12`

When the rocket completes its trajectory and hits the ground the time t can not be 0. When t =0, the rocket starts the trajectory.

So when the rocket completes its trajectory and hits the ground ,

then t = 12seconds.

So we have got the required answers.