Question

A rocket is launched from a tower. The height of the rocket, y in feet, isrelated to the time after launch, x in seconds, by the given equation. Usingthis equation, find the time that the rocket will hit the ground, to the nearest100th of second.Y= -16x^2 + 121x + 83

Answer 1

To solve this, we must solve the equation given.

As when Y = 0, the rocket hits the ground,

-16x² + 121x +83 = 0

using bhaskara:

Delta = b² -4ac

= 121² -4(-16)(83)

= 14641 +5312

= 19953

x1= (-b + sqrt19953)/2a

and

x2 = (-b - sqrt19953)/2a

x1= (-121 + 141,25)/-32

x1 = -0.63 s

x2= (-121 - 141,25)/-32

x2 = 8.195 s

As there's no such thing as negative time, the answer is x2 = 8.195s. Ronded to the nearest 100th, the answer is 8.20 seconds.