Question

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=16t^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 the time when the rocket hits the ground, we need to set h equal to 0 and solve for t.

We have the equation:

0 = 16t^2 + 113t + 65

This is a quadratic equation, so we can use the quadratic formula to find the solutions. The quadratic formula is:

t = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 16, b = 113, and c = 65. Substituting these values into the quadratic formula, we get:

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

Simplifying the expression inside the square root gives us:

t = (-113 ± √(12769 - 4160)) / 32

t = (-113 ± √(8609)) / 32

t = (-113 ± 92.79) / 32

The two possible solutions are:

t = (-113 + 92.79) / 32 = -20.21 / 32 ≈ -0.63s

t = (-113 - 92.79) / 32 = -205.79 / 32 ≈ -6.43s

Since we're only interested in positive times, we can ignore the negative solution. Therefore, the rocket will hit the ground approximately 6.4 seconds after it is launched.