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.
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.