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+103t+290

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+103t+290

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 ____   meters above sea-level.

Answer 1

h(t) = -4.9t^2 + 103t +290

take the

h'(t) = -9.8t + 103 = 0

max height is reached at time t=103/9.8 = about 10.5 seconds

max height = h(10.5) = -4.9(10.5)^2 +103(10.5) + 290 = -540.225 + 1081.5= about 831 meters

splashdown is found by setting h(t) =0

h(t) -4.9t^2 + 103t + 290 = 0

use the quadratic formula. You should get about twice the time for max height or about 21+ seconds

t= 103/9.8 + (1/9.8)sqr(103^2 +4(4.9)(290)) = 10.5 + 13 = 23.5 seconds to splashdown