Word problem (pre calc)

Question

NASA launches a rocket t=0 seconds. It’s height , in meters above sea level-, as a function of time is given by h(t)=-4.9t^2+298t+353

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

how high above sea level does the rocket get at its peak?

Answer 1

Question 1: Assuming that the rocket will splash down into the ocean, at what time does splashdown occur?

We can find the answer to this by finding when the height is at sea level, or h(t) = 0. This should happen after t=0, since the rocket splashes down into the ocean after launch.

h(t) = -4.9t² + 298t + 353 = 0

Using the quadratic formula to solve for the roots:

t = [-298 ±√(298² - 4*-4.9*353)]/(2*-4.9)

t = 61.97 seconds (disregard negative solution)

Therefore, the rocket splashes down after 61.97 seconds.

Question 2: How high above sea-level does the rocket get at its peak?

The maximum of the second order polynomial occurs when t = -b/(2a)

t = -298/(2*-4.9) = 30.41 seconds

Plugging this time back into the height equation:

f(30.41) = -4.9(30.41))² + 298(30.41) + 353 = 4883.82 meters

Therefore, the rocket peaks at 4883.82 meters above sea-level.