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.