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Quadratic Equation HELP ASAP?

A tennis ball is hit from an initial height of 0.5m. The equation that models the path of the tennis ball is h(t)= -0.125d^2 + d + 0.5 , where h is the height of the ball, in meters, and d is the horizontal distance from the tennis player, in meters.

A) Find the maximum height of the ball and the distance from the tennis player when the ball reaches maximum height, both in meters.

B) How far has the tennis ball travelled horizontally, in meters, when it hits the ground?

C) For what horizontal distance, in meters, is the tennis ball higher than 2m?

D) The ball was hit directly toward the net, which is 0.6m from the player and 0.95 min height. Will the ball make it over the net? Justify your answer.

1 Answer

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  • ?
    Lv 7
    1 month ago

    Why "h(t)" ?  The time is not a variable in this problem.  But I'm going to call the horizontal distance by the name "x" rather than "d", so that I can write a derivative without confusion.

    (a)  h(x) = -0.125x^2 + x + 0.5.

    dh/dx = -0.250x + 1 => max height reached when 1 = 0.250x => x = 4 m.

    (b)  h=0 => 0 = 0.125x^2 - x - 0.5 =>

    0 = x^2 - 8x - 4 =>

    x = 4 + (1/2)*sqrt(64+16) 

    = 4 + 2*sqrt(5) = around 8.47 m.

    (c)  When h = 2m, you have

    0 = -0.125x^2 + x - 1.5 =>

    0 = x^2 - 8x + 12 => 

    x = 2 or 6, so there is a 4-meter interval where the ball is higher than 2m.

    (d)  h(0.6 m) = -0.125*(0.6)^2 + 0.6 + 0.5

    = 1.1 m - 0.36 m/8 which is more than 1m, so high enough.

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