When the skater starts 7 mm above the ground, how does the speed of the skater at the bottom of the track compare to the speed of the skater

Question

When the skater starts 7 mm above the ground, how does the speed of the skater at the bottom of the track compare to the speed of the skater at the bottom when the skater starts 4 mm above the ground?

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Kim Cúc 4 years 2021-08-24T16:25:16+00:00 1 Answers 39 views 0

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  1. kimchi2
    0
    2021-08-24T16:26:39+00:00
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    Answer:

    Speed is higher and 1.32 times greater.

    Explanation:

    Considering downward motion as positive.

    Given:

    Case 1:

    Initial height (h₁) = 7 mm = 0.007 m [1 mm =0.001 m]

    Acceleration due to gravity (g) = 9.8 m/s²

    Initial velocity (u₁) = 0 m/s

    Final velocity (v₁) = ?

    Using conservation of energy, we have:

    Decrease in potential energy = Increase in Kinetic energy

    mgh_1=\frac{1}{2}mv_1^2\\\\v_1=\sqrt{2gh_1}=\sqrt{2\times 9.8\times 0.007}=0.37\ m/s

    Case 2:

    Initial height (h₂) = 4 mm = 0.004 m

    Acceleration due to gravity (g) = 9.8 m/s²

    Initial velocity (u₂) = 0 m/s

    Final velocity (v₂) = ?

    Using conservation of energy, we have:

    Decrease in potential energy = Increase in Kinetic energy

    mgh_2=\frac{1}{2}mv_2^2\\\\v_2=\sqrt{2gh_2}=\sqrt{2\times 9.8\times 0.004}=0.28\ m/s

    From the above values, we can conclude:

    v_1>v_2

    Also,

    \frac{v_1}{v_2}=\frac{0.37}{0.28}=1.32\\\\v_1=1.32v_2

    So, the velocity in the first case is 1.32 times greater than velocity in second case.

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