NASA launches a rocket at t=0

Question

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+124t+416. Assuming that the rocket will splash down into the ocean, at what time does splashdown occur?The rocket splashes down after     seconds.How high above sea-level does the rocket get at its peak?The rocket peaks at 

Answer 1

h(t) = −4.9t² + 124t + 416

Splash down means h(t_splash) = 0, we need to find the roots of the polynomial.

−4.9t_splash² + 124t_splash + 416 = 0

when we multiply both sides of the equation by 10

−49t_splash² + 1240t_splash + 4160 = 0

t_1,2 = [-b +- sqrt(b^2 - 4ac)] / 2a

for at² + bt + c = 0

the positive root value gives us the splash down time

The peak height occurs for the t value at h'(t) = 0 which is

h'(t_peak) = -9.8t_peak + 124 = 0

You can calculate the t_peak from above and put in h(t) to find the peak distance