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+43t+339

.


(A) Assuming that the rocket will splash down into the ocean, at what time does splashdown occur? (Round answer to 2 decimal places)






The rocket splashes down after

seconds._______






(B) How high above sea-level does the rocket get at its peak? (Round answer to 2 decimal places)






The rocket peaks at _____

meters above sea-level._____

Answer 1

(A)

`- 4.9 {t}^(2) + 43t + 339 = 0`

`49 {t}^(2) - 430t - 3390 = 0`

`t = \frac{ - ( - 430) + \sqrt{ {( - 430)}^(2) - 4(49)( - 3390)} }{2(49)} = (430 + √(849340) )/(98) = 13.79`

The rocket splashes down after 13.79 seconds.

(B) h'(t) = -9.8t + 43 = 0

t = 43/9.8 = 215/49 = 4.39 seconds

h(4.39) = 433.34 meters

At t = 4.39 seconds, the rocket peaks at

433.34 meters above sea level.