h(t)=-4.9t^2 + 100t+206

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)=−4.9t^2 +100t+206  .

A) Assuming that the rocket will splash down into the ocean, at what time does splashdown occur?

The rocket splashes down after ________________ seconds.

 

B)How high above sea-level does the rocket get at its peak?
The rocket peaks at _______________meters above sea-level.

Answer 1

-4.9t^2 + 100t + 201

take the derivative and set = 0

-9.8t + 100 = 0

maximum height is reached when t =100/9.8 = about 10.2 seconds

plug that into the original equation to solve for maximum height

-4.9(100/.98)^2 + 100(100/9.8) + 201

= -510.2 + 1020.4 + 201 = 711.2 meters = maximum peak height

for splashdown

0 = -4.9t^2 +100t + 201

solve for t,

use the quadratic formula

t =100/9.8 + or - (1/9.8)sqr(100^2 +4(4.9)201)) =about 22.25 seconds to splashdown