# What is the answer to the following problem in projectile motion:Two physics students built a model rocket for a class project. Once the rocket is launched, its height in feet, h(t), can be found...

What is the answer to the following problem in projectile motion:

Two physics students built a model rocket for a class project. Once the rocket is launched, its height in feet, h(t), can be found using the function h(t) = -16t² + 96t, where t represents time in seconds. One of the students records the time from the moment the rocket launches until it falls to the ground.

(A) Find the height of the rocket at 5 seconds. Explain how you solved this problem. Provide complete details and answer in a complete sentence.

(B) At what other time will the rocket be at the same height as your answer form part A? Explain how you found the answer to this question.

(C) Explain why we have two different values for the same height in parts A and B.

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Two physics students built a model rocket for a class project. Once the rocket is launched, its height in feet, h(t), can be found using the function h(t) = -16t² + 96t, where t represents time in seconds.

The height if the rocket at t = 5 is given by h(5) = -16*5^2 + 96*5 = -16*25 + 96*5 = 80.

The height of the rocket is equal to 80 at the time t where. Solving the quadratic equation gives:

16*t^2 - 96t = -80

=> t^2 - 6t + 5 = 0

=> t^2 - 5t - t + 5 = 0

=> t(t - 5) - 1(t - 5) = 0

=> (t - 1)(t - 5) = 0

=> t = 1 and t = 5

The rocket is at a height of 80 at t = 1 and t = 5

The path followed by the rocket is that of a parabola. It is at a height of 80 first while going up and then again while coming down.

Part A

h(5) = -16(5)^2 +96(5)

h(5) = 80 feet at t = 5 seconds

Part B

80 = -16t^2 + 96t

16t^2 - 96t + 80 = 0

t^2 - 6t + 5 = 0

(t - 5) (t - 1) = 0

The height of the rocket is 80 feet at t = {1 second or 5 seconds)

Part C

At one second, the rocket is 80 feet high on the way up. At five seconds the rocket is 80 feet high on the way down.