How do I solve this problem?

Question

The path of a rocket fired during a fireworks display is given by the equation s(t) = 64t^2− 16t, where t is the time, in seconds, and s is the height, in feet.

What is the maximum height, in feet, the rocket will reach?

In how many seconds will the rocket hit the ground?

Answer 1

Actually, there's a typo in the question, and none of the below is valid. Somebody missed a minus sign. As matters stand with the given equation, the rocket starts off by burrowing into the ground, then blasts back out and keeps going infinitely far up into the sky. :)

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To get the number of seconds, note that you're being asked about the root of s(t) = 0 other than t = 0. I.e., s(t) = 0 when t = 0 and then again when t = 1/4, and so the whole thing is over in a quarter-second.

If you know calculus, take the first derivative of s and set it equal to 0, and you'll be at a maximum.

If not, and you know about parabolas, all is still cool.

If neither, assert that it's "obvious" due to "symmetry" that the maximum must be reached at the halfway point, namely when t = 1/8.

Answer 2

I suspect the equation should be:
s(t) = 64t - 16t^2
so that the rocket goes up and then comes down. This is also consistent with a value of 32ft/sec^2 for the acceleration due to gravity.

The rocket reaches maximum height when:
s'(t) = 0
64 - 32t = 0
t = 2sec.

Substituting that value of t in (1) gives:
s(2) = 128 - 64 = 64ft.

The rocket is at ground level when s(t) = 0:
t(64 - 16t) = 0
t = 0sec. (take-off)
t = 4sec. (landing).