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?

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  1. 8 feet away horizontally, vertically, or line-of-sight?

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  2. 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.

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