The height, in meters, above sea level of a NASA launched rocket is given by h ( t ) = − 4.9 t^2 + 301 t + 94

Question

The height, in meters, above sea level of a NASA launched rocket is given by h(t)=−4.9t2+301t+94 where t represents the number of seconds since the rocket was launched.

Assuming that the rocket will splash down into the ocean, after how many seconds will the splashdown occur?

How many meters above sea level does the rocket get at its peak?

If we assume that the rocket is launched from a launch pad, how high is the launch pad?

Answer 1

h(t)=-4.9t^2+301t +94

height of launching pad = 94 meters

v(t)= h'(t) =-9.8t+301= 0

t= 301/9.8 = about 30.71 seconds to reach max height

peak height = -4.9(301/9.8)^2 +301(301/9.8)+ 94

=about 4,716.5 meters

time to splash down is a little more than twice time to peak height

= 61.739 seconds

set h(t)=0 and solve for t

-4.9t^2+301t+94=0

use the quadratic formula, ignore the negative square root