A model rocket is launched from a raised platform.

Question

Its height in feet is given by H = -16t² + 160t + 20 (t = seconds after launch)

After how many seconds will the rocket be 116 ft high?

When will the rocket hit the ground?

Answer 1

Does 20 represent the platform? If so:

t=.6415sec

At 116' the ground will be hit at (t^2+t)/2*32.2=2.2305sec

Answer 2

116 = - 16t² + 160t + 20, 16t² - 160t + 96 = 0; t² - 10t + 6 = 0. t = 5 ± √19;

t = .6411 s going up, t = 9.3589 s going down

H = 0, 16t² - 160t - 20 = 0; 4t² - 40t - 5 = 0, t² - 10t - 1.25 = 0, t = 5 ± √26.25, t = 5 + √26.25 = 10.1235 s

object will land at t = 10.1235 s

Answer 3

After how many seconds will the rocket be 116 ft high?

H = 116ft = -16t² + 160t + 20

0 = -16t² + 160t - 96

0 = -16(t² - 10t + 6)

0 = t² - 10t + 6

t = [10 ± √((-10)² - 4(1)(6))] / 2(1) = [10 ± √(76)] / 2 = 0.6411 or 9.3589

The first time the rocket hits 116ft will be after 0.6411 seconds.

When will the rocket hit the ground?

H = 0ft = -16t² + 160t + 20

t = [-160 ± √((160)² - 4(-16)(20))] / 2(-16) = [-160 ± √(26880)] / (-36) = -0.1098 or 8.9986

We reject the negative answer, since time can't be negative. The rocket will hit the ground after 8.9986 seconds.