How do you find the maximum height and time of flight for the rocket in question 32?

s(t) = 64t − 16t²

This can be answered by completing the square:

64t − 16t² = -16[t² - 4t)

= -16[(t - 2)² - 4]

To maximise this function, we plainly have to minimise the bit in the square brackets. The minimum value of (t - 2)² is zero, when t = 2. That gives

s_max = -16[0 - 4] = -16 x -4 = 64 feet.

Alternatively, we can use differentiation to find the vertical velocity, v.

s(t) = 64t − 16t²

Therefore v(t) = 64 - 32t. The value of v is zero as the rocket attains maximum height, so solve

64 - 32t = 0

t = 2. Substitute this into the expression for s and we will again find that the maximum height is 64 feet...which is fairly feeble come to think of it.

In how many seconds will the rocket hit the ground?

This is simply a case of solving s(t) = 0.

64t − 16t² = 0

Factorise: t(64 - 16t) = 0

So either t = 0 (i.e. at launch, which does not interest us.)

Or (64 - 16t) = 0 and therefore t = 4 seconds. Note that, because the rocket's trajectory describes a symmetrical parabolic path, we could simply have doubled the time taken to reach maximum height: 2 x 2 = 4. The rocket takes the same amount of time to come down as it did to go up.

First answer by Plenilune. Last edit by Plenilune. Contributor trust: 122 [recommend contributor]. Question popularity: 6 [recommend question]