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.
Answers
GPT 3.5
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.
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.