A cliff jumper jumps off a cliff with an initial horizontal velocity of 10 m/s. The cliff is 10 meters high. How far from the base of the cl

Question

A cliff jumper jumps off a cliff with an initial horizontal velocity of 10 m/s. The cliff is 10 meters high. How far from the base of the cliff does the diamond fall?

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Nick 4 years 2021-07-28T20:32:44+00:00 1 Answers 15 views 0

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  1. vankhanh
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    2021-07-28T20:33:50+00:00
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    Answer:

    The distance reached is 14.3 m.

    Explanation:

    To find the distance reached by the diamond first we need to find the flight time:

    t_{v} = \sqrt{\frac{2h}{g}}

    Where:

    h: is the height = 10m

    g: si the gravity = 9.81 m/s²

    t_{v} = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2*10 m}{9.81 m/s^{2}}} = 1.43 s

    Now, we can find the distance reached:

    x = V_{0_{x}}*t_{v}

    Where:

    V_{0_{x}} is the initial horizontal velocity = 10 m/s

    x = V_{0_{x}}*t_{v} = 10 m/s*1.43 s = 14.3 m

    Therefore, the distance reached is 14.3 m.

    I hope it helps you!    

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