Arocket 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.

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

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.

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

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

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

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.

So by using zero product property we will get

Also

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.

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

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

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

ans also

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.

mathpapa

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