Question

A rocket is launched from atop a 101 foot cliff with an initial velocity of 116 feet per second. After how many seconds will the rocket land? Round to the nearest second.

Answer 1

8 seconds

Explanation:

Assuming a vertical launch, the position of the rocket is given by

xf = xo + vo t + 1/2 at^2

xo = 101 ft vo = 116 xf = 0 (when it hits the ground) a = -32.2 ft/s^2

0 = 101 + 116 t - 16.1 t^2 Use Quadratic Formula to find t = ~~ 8 sec