A high diver of mass 60.0 kg steps off a board 10.0 m above the water and falls vertical to the water, starting from rest. If her downward m

Question

A high diver of mass 60.0 kg steps off a board 10.0 m above the water and falls vertical to the water, starting from rest. If her downward motion is stopped 2.10 s after her feet first touch the water, what average upward force did the water exert on her

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Thạch Thảo 4 years 2021-08-26T08:43:14+00:00 1 Answers 4 views 0

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  1. thienan
    0
    2021-08-26T08:44:47+00:00
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    Answer:

    The average upward force exerted by the water is 988.2 N

    Explanation:

    Given;

    mass of the diver, m = 60 kg

    height of the board above the water, h = 10 m

    time when her feet touched the water, t = 2.10 s

    The final velocity of the diver, when she is under the influence of acceleration of free  fall.

    V² = U² + 2gh

    where;

    V is the final velocity

    U is the initial velocity = 0

    g is acceleration due gravity

    h is the height of fall

    V² = U² + 2gh

    V² = 0 + 2 x 9.8 x 10

    V² = 196

    V = √196

    V = 14 m/s

    Acceleration of the diver during 2.10 s before her feet touched the water.

    14 m/s is her initial velocity at this sage,

    her final velocity at this stage is zero (0)

    V = U + at

    0 = 14 + 2.1(a)

    2.1a = -14

    a = -14 / 2.1

    a = -6.67 m/s²

    The average upward force exerted by the water;

    F_{on\ diver} = mg - F_{ \ water}\\\\ma = mg - F_{ \ water}\\\\F_{ \ water} = mg - ma\\\\F_{ \ water} = m(g-a)\\\\F_{ \ water} = 60[9.8-(-6.67)]\\\\F_{ \ water} = 60 (9.8+6.67)\\\\F_{ \ water} = 60(16.47)\\\\F_{ \ water} = 988.2 \ N

    Therefore, the average upward force exerted by the water is 988.2 N

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )