A model rocket is launched upward with an initial velocity o

A model rocket is launched upward with an initial velocity of 200 feet per second. The height, in feet, of the rocket t seconds after the launch is given by h = -16t^2 + 1200t. How many seconds after the launch will the rocket be 350 feet above the ground? Round to the nearest tenth of a second. (Enter your answers as a comma-separated list.)

Solution

h(t) = -16t^2 + 200

If h(t) = 350 feet

-16t^2 + 200t = 350

solve for t : -16t^2 +200t -350 =0

use quadratic root formula:

t = ( -200 +/- sqrt(200^2 -4*350*16))/-32

= 2.104 , 10.396

So, t = 2.1 sec , 10.4 sec