Two physics students shoot a water-bottle rocket from a 1 m tall stand. They calculate that rocket will travel with the given formula, f(x)=–16t 2+64t+1 where f is its height in feet above ground t seconds after being released. After how many seconds will the rocket reach its maximum height? What is that height?

2 secs; 65 feet

Explanation:

The function guiding the water bottle is given as:

f(x) = -16t² + 64t + 1

The bottle will reach maximum height when velocity, df/dt (velocity is the first derivative of distance) = 0.

Therefore:

df/dt = 0 = -32t + 64

=> 32t = 64

t = 64/32 = 2 seconds

This is the time it will take to reach the maximum height.

To find this height, we insert t = 2 into the function f(x):

f = -16(2)² + 64(2) + 1

f = -(16 * 4) + 128 + 1

f = -64 + 128 + 1

f = 65 ft

Its maximum height is 65 ft.

It takes the water bottle rocket 2 s to reach maximum height.

The maximum height reached is 65 m

Explanation:

The object would reach maximum height when its velocity is zero. That is f'(x)= 0. So, we differentiate f(x) to get

df(x)/dx = d(-16t² + 64t + 1)/dt = -32t + 64 = 0

-32t + 64 = 0

-32t = -64

t = -64/-32 = 2 s

So it takes the water bottle rocket 2 s to reach maximum height.

To find this height, we substitute t = 2 into f(x). So,

f(x) = -16t² + 64t + 1

f(2) = -16(2)² + 64(2) + 1 = -16(4) + 128 + 1 = -64 + 128 + 1 = 65 m

So the maximum height reached is 65 m.

the answer would be c. motion and direction

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