NASA launches a rocket at t = 0 seconds. Its height, in meters above sea-level, as a function of time is given by h(t) = − 4.9t² + 40t + 262.

Question

Assuming that the rocket will splash down into the ocean, at what time does splashdown occur? I The rocket splashes down after seconds. How high above sea-level does the rocket get at its peak? The rocket peaks at meters above sea-level.

Answer 1

Ashley,

Can you follow the method which I used to solve the other problem for you? With the child throwing the ball? I will check back tomorrow and see what you have done and see if you have any questions.

(It is possible another Tutor will jump in and do it for you. I have no control over that.)

Answer 2

h(t)=-4.9t^2 + 40t + 262

take the derivative and set = 0 to find time when it reaches max height

h'(t) = -9.8t +40 = 0

t = 40/9.8 = 20/4.9= about 4.08 seconds

plug that t value back into the original equation to solve for max height

h(20/4.9) = -4.9(20/4.9)^2 +40(20/4.9) + 262 = (20/4.9)[-20+40] + 269 = 400/4.9+262

= about 343.63 meters = max height