A model rocket is launched

Question

A model rocket is launched with an initial velocity of 150 ft/s. The height, h, in feet, of the rocket t seconds after the launch is given by h = −16t² + 150t.

 How many seconds after launch will the rocket be 260 ft above the ground? Round to the nearest hundredth of a second.

Answer 1

h(t) = -16t² + 150t, where t is time (in seconds) and h is the height (in feet).

We want to find what time will the rocket reached 260 feet above ground after the launch. We will rewrite this in standard form and solve for t.

-16t² + 150t = 260

-16t² + 150t - 260 = 0

-2(8t² - 75t - 130) = 0

Use the quadratic formula with a = 8, b = -75, and c = -130.

x = [-b ± √(b² - 4ac)] / 2a

x = [-(-75) ± √((-75)² - 4(8)(-130))] / 2(8)

x = [75 ± √(5625 + 4160)] / 16

x = [75 ± √(9785)] / 16

x ≈ -1.49 or x ≈ 10.87 (negative solution will not work in this problem)

The rocket will reach the height of 260 feet above ground after 10.87 seconds.