The height h in feet of an object after t seconds is given by the function h=-16t^2+60t+9....
How long will it take the object to hit the ground? Round your answer to the nearest thousandth....
when it hits the ground, h is zero. So you have a quadratic equation. Use the quadratic equation to solve...
-16t^2+60t+9=0
The height h in feet of an object after t seconds is given by the function h=-16t^2+60t+9....
How long will it take the object to hit the ground? Round your answer to the nearest thousandth
The formula you give is for determining the height reached by an object when launched from a height of 9 feet above the ground, h = 9 + Vot - gt^2 where h == the maximum height, Vo = the initial velocity and g = the acceleration due to gravity, 32 fps^2.
Your formule implies a Vo of 60 fps.
The time to reach the maximum height derives from Vf = Vo - gt or 0 = 60 - 32t making t = 1.875 sec.
The height reached is h = 9 + 60(1.875) - 16(1.875^2) = 65.25 ft.
The time to fall back down to the ground, 9 feet below the launch height of 9 feet, derives from
74.25 = 0 + 16t^2 making t = 2.15 sec.
The total time of flight is therefore 4.029 sec.
To solve for the time it takes for the object to hit the ground, we need to find the value of t when the height h is equal to 0.
Using the given function, h = -16t^2 + 60t + 9, we can set it equal to 0:
-16t^2 + 60t + 9 = 0
This is a quadratic equation, and we can solve it using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = -16, b = 60, and c = 9. Plugging in these values, we get:
t = (-60 ± √(60^2 - 4(-16)(9))) / (2(-16))
Simplifying further, we have:
t = (-60 ± √(3600 + 576)) / -32
t = (-60 ± √4176) / -32
Now, taking the square root of 4176, we get:
√4176 ≈ 64.637
Now we can substitute this value into the equation:
t = (-60 ± 64.637) / -32
Now we can calculate the two possible values for t:
t1 = (-60 + 64.637) / -32 ≈ 0.133
t2 = (-60 - 64.637) / -32 ≈ 3.766
We have two possible values for t, but we are only interested in the time it takes for the object to hit the ground. Therefore, we can ignore the positive value t1 = 0.133.
Hence, the object will hit the ground approximately 3.766 seconds after it was launched.