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31 August, 16:08

How far would the ball be above a fence 3.0 m (10 ft) high if the fence was 116 m (380 ft) from home plate?

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  1. 31 August, 16:34
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    Let V be the initial speed of the ball. Because the ball was hit angle an angle, there are two speed components: one is in x-direction, and the other one is in the y-inderect. The ball travels at a constant speed horizonally but accelerates vertically.

    Break down the intial speed

    Vx = Vcos45

    Vy = Vsin45

    find the time it takes the ball to hit the ground 116m away horizonally

    x = vt

    116 = Vcos45 * t

    t = 116 / Vcos45

    Xf =.5at^2 + Vt + Xi

    Xf = final position (0m)

    Xi = initial position (.9m)

    a = acceleration

    t = time

    V = speed in y-derection

    we just found the time, which is 188 / Vcos45, plug it in

    0 =.5 (-9.8) (188 / Vcos45) ^2 + Vsin45 * (188 / Vcos45) +.9

    -.9 = - 4.9 (188 / Vcos45) ^2 + 188

    -188.9 = - 4.9 (188 / Vcos45) ^2

    38.551 = (188 / Vcos45) ^2

    6.208 = 188 / Vcos45

    6.208 * Vcos45 = 188

    V = 188 / (6.208cos45)

    V = 42.82 m/s

    The ball was hit with the speed of 42.82m/s

    Find the it takes the ball to travel 116m

    x = vt

    116 = 42.82cos45 * t

    t = 3.831s

    Xf =.5at^2 + Vt + Xi

    Xf =.5 (-9.8) (3.831) ^2 + 42.82sin45 (3.831) +.9

    Xf = 45m

    the ball was 45 - 3 = 42m above the fence
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