the height of a rocket after being launched is modeled by h(t)=-16t^2+200t at what time will the rocket reach its maximum height?\

Answer 1

This looks like a calculus problem viewing physics rather than a physics problem using calculus: not that it really matters I was just saying.

Anyhow, the trick here is a key interpretation idea. When will the rocket reach it's max. height?

When it's velocity equals zero!

h(t)=-16t^2+200t

Take the derivative of this guy (displacement) to find instantaneous velocity or v(t).

h´(x) = v(t) = -32t + 200

Set this equal to 0.

0 = -32t + 200 = -8(4t - 25)

t = 25/4

Therefore, in 6.25 seconds the rocket will achieve maximum height.

And, if you need to find that height use t = 6.25s in the original equation.

Answer 2

Hello Rialey! You can solve this problem different ways depending on your level of math. The most direct method is using differential calculus. Take the derivative of the function with respect to t and then set it equal to 0. Solving for t will give you the time for maximum height. The derivative calculates the slope of the function at any point so by setting it to zero we are solving for the maximum in the function.

dh/dt = -32t +200 = 0

Solving for t gives 6.25 seconds at maximum height.

If you have a graphing calculator you can also solve if by graphing the function and then solve for the time at the peak.