A rocket is fired upward from initial distance

Question

A rocket is fired upward from some initial distance above the ground. It’s height in feet (h) above the ground seconds (t) after it is fired, is given by h(t)=-16t^2+192t+208

what is the Rockets maximum height?

How long does it take for the rocket to reach its maximum height?

Answer 1

h(t) = -16t^2 + 192t + 208

take the derivative with respect to time t, then set it equal to zero

velocity = h'(t) =-32t + 192 = 0 when at a maximum height

solve for t

-32t = -192

t = 192/32 = 6 seconds to reach max height

plug that into the original formula to solve for max height = h(6)

= -16(6^2) + 192(6) + 208

=-16(36) + 1152 + 208

= -576 +1156 + 208

= 576 +208

= 784 feet = maximum height reached in 6 seconds from liftoff

that's the calculus method

you could also solve with algebra and geometry. Plot the parabola and find the vertex = (h,k) = (784, 6)

use a handheld or online graphing calculator, such as desmos

or

rewrite the formula in vertex form

h(t) = -16t^2 + 192t +208

= -16(t^2 - 12t + 36) + 208 +16(36)

= -16(t-6)^2 + 208+ 576

= -16(t-6)^2 + 784

= -16(t-h)^2 + k where (h,k) = (6,784) and h=6 seconds and k=784 feet = max height

3 solution methods, calculus, geometry, algebra