A cliff diver from the top of a 100 (m) cliff. He begins his dive by jumping up with a velocity of 5 (m/s) a. How long does it take fo

Question

A cliff diver from the top of a 100 (m) cliff. He begins his dive by jumping up with a velocity of 5 (m/s) a. How
long does it take for him to hit the water below? b. What is his velocity right before he hits the water?

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Linh Đan 2 months 2021-07-23T11:31:21+00:00 1 Answers 3 views 0

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    2021-07-23T11:32:57+00:00

    Answer:

    A.) 4.81 seconds

    B.) 44.6 m/s

    Explanation:

    He begins his dive by jumping up with a velocity of 5 (m/s).

    Let us first calculate the maximum height reached by using third equation of motion

    V^2 = U^2 – 2gH

    At maximum height, V = 0

    0 = 5^5 – 2 × 9.8H

    19.6H = 25

    H = 25 /19.6

    H = 1.28 m

    The time taken for the diver to reach the water from the maximum height can be calculated by using second equation of motion.

    Where height h = 1.28 + 100 = 101.28 m

    h = Ut + 1/2gt^2

    As the diver drop from maximum height, U = 0

    101.28 = 1/2 × 9.8 × t^2

    4.9t^2 = 101.28

    t^2 = 101.28/4.9

    t^2 = 20.669

    t = sqrt ( 20.669)

    t = 4.55s

    As the diver jumped up, the time taken to reach the maximum height will be

    Time = 1.28 / 5 = 0.256

    The time taken for him to hit the water below will be 0.256 + 4.55 = 4.81 seconds

    B.) Velocity right before he hits the water will be

    V^2 = U^2 + 2gH

    But U = 0

    V^2 = 2 × 9.8 × 101.28

    V^2 = 1985.09

    V = 44.6 m/s

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