A rocket is launched from atop a 65-foot cliff with an initial velocity of 113 ft/s. The height of the rocket above the ground at time t is given by h=-16x^2+113t+65 . When will the rocket hit the ground after it is launched? Round to the nearest tenth of a second.

Answer 1

To find when the rocket hits the ground, we need to determine the value of t when h=0.

Substituting h=0 into the equation gives us:
0=-16t^2+113t+65

To solve this equation, we can factor it or use the quadratic formula. In this case, we will use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / 2a

For our equation, a=-16, b=113, and c=65.

t = (-113 ± √(113^2 - 4*(-16)*65)) / (2*(-16))

Simplifying further:

t = (-113 ± √(12769 + 4160)) / (-32)
t = (-113 ± √(16929)) / (-32)

Finding the square root value:

t = (-113 ± 130) / (-32)

Now, calculating both possible values of t:

t1 = (-113 + 130) / (-32) = 17 / -32 ≈ -0.5
t2 = (-113 - 130) / (-32) = -243 / -32 ≈ 7.6

The rocket will hit the ground approximately 7.6 seconds after it is launched.