A ball is kicked off the top of a cliff that is 30 feet tall at an angle of 45° to the horizontal with an initial velocity of 25 feet per second. The quadratic equation shown below models the height, h(x), of the ball when it is x feet from the cliff’s edge. How high above the ground will the ball be when it is 8 feet from the cliff’s edge?
The answers are:
22.73 feet
34.72 feet
41.28 feet
53.89 feet
I got 53.89 is it right?
8 feet away horizontally, vertically, or line-of-sight?
Vo = 25ft/s[45o].
Xo = 25*Cos45 = 17.7 ft/s.
Yo = 25*sin45 = 17.7 ft/s.
Xo * T = 8.
17.7T = 8,
T = 0.452 s. = time in air.
h = ho + Yo*T - 16*T^2.
h = 30 + 17.7*0.452 - 16*(0.452)^2 = 34.73 ft.
To find the height of the ball when it is 8 feet from the cliff's edge, we need to substitute x = 8 into the quadratic equation for h(x).
First, let's rewrite the given quadratic equation for the height h(x) of the ball as a function of x:
h(x) = -0.0026x^2 + 0.2174x + 30
Now, to find the height when x = 8, substitute x = 8 into the equation:
h(8) = -0.0026(8)^2 + 0.2174(8) + 30
Calculating this expression:
h(8) = -0.0026(64) + 0.2174(8) + 30
h(8) = -0.1664 + 1.7392 + 30
h(8) = 1.5728 + 30
h(8) ≈ 31.5728
Therefore, the ball will be approximately 31.5728 feet above the ground when it is 8 feet from the cliff's edge.
None of the given answer options are close to 31.57 feet, so it appears that 53.89 feet is not the correct answer.