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18 February, 11:52

A long jumper jumps at a 20-degree angle and attains a maximum altitude of 0.6 m. What is her initial speed? [10m/s] How far is her jump? [6.59m]

I need someone to explain to me how to get the answers listed above my showing a detailed step to step process.

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  1. 18 February, 12:00
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    Refer to the diagram shown below.

    Assume g = 9.8 m/s² and ignore air resistance.

    Let V = the initial speed

    Let d = the distance raveled (how far she jumps).

    The horizontal velocity is

    Vx = V cos (20°) = 0.9397V m/s

    The vertical launch velocity is

    Vy = V sin (20°) = 0.342V m/s

    At maximum height of 0.6 m, the vertical velocity is zero. Therefore

    Vy² - 2 * (9.8 m/s²) * (0.6 m) = 0

    (0.342V m/s) ² = 11.76 (m/s) ²

    0.342V = 3.4293

    V = 10.027 m/s

    Therefore

    Vy = 0.342*10.027 = 3.4239 m/s

    Vx = 0.9397*10.027 = 9.4226 m/s

    The time to attain maximum height is a half of the total time of travel. This time is given by

    Vy - gt = 0

    3.4239 = 9.8*t

    t = 3.4239/9.8 = 0.3494 s

    The total travel time is 2*0.3494 = 0.6988 s

    The length of the jump is

    Vx * (2t) = (9.4226 m/s) * (0.6988 s) = 6.5845 m

    Answer:

    The initial speed is 10.0 m/s (nearest tenth)

    The length of the jump is 6.58 m (nearest hundredth)
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