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23 April, 19:53

A ball is thrown straight up with an initial velocity of 50 m/s.

a) How high will it go?

b) How long does it take to reach this height?

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  1. 23 April, 20:15
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    a.) Maximum height = 127.4 metre

    b.) Time = 5.1 seconds

    Explanation:

    Given that initial velocity U = 50m/s

    a.) How high will it go means maximum height.

    At maximum height, final velocity V = 0

    Using the equation 3 of linear motion

    V^2 = U^2 - 2gH

    As the ball is going up, g will be negative.

    0 = 50^2 - 2 * 9.81H

    2500 = 19.62H

    H = 2500/19.62

    H = 127.4 meters

    b.) The time taken to reach the maximum height

    Using the 2nd equation of motion

    h = Ut + 1/2gt^2

    127.4 = 50t - 1/2 * 9.8t^2

    127.4 = 50t - 4.9t^2

    4.9t^2 - 50t + 127.4 = 0

    This will lead to quadratic equation

    Where a = 4.9, b = - 50, c = 127.4

    Using quadratic formula will lead to

    50 + / - root (50^2 - 4 * 4.9 * 127.4) / 2 * 4.9

    50 + / - root (2500 - 2497.04) / 9.8

    50+ / - root (2.96) / 9.8

    50+ / - 1.72/9.8

    (50 + 1.72) / 9.8 or (50 - 1.72) / 9.8

    51.72/9.8 = 5.28s

    Or 48.28/9.8 = 4.9s

    To test for this let's assume it's a perfect parabola. Using vertex formula,

    t = - b/2a

    t = 50/9.8

    t = 5.1 m/s
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