A 0.26-kgkg rock is thrown vertically upward from the top of a cliff that is 32 mm high. When it hits the ground at the base of the cliff, t

Question

A 0.26-kgkg rock is thrown vertically upward from the top of a cliff that is 32 mm high. When it hits the ground at the base of the cliff, the rock has a speed of 33 m/sm/s .

a. Assuming that air resistance can be ignored, what is the initial speed of the rock?
b. What is the greatest height of the rock as measured from the base of the cliff?

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Ladonna 1 year 2021-08-15T01:48:47+00:00 1 Answers 14 views 0

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    2021-08-15T01:50:08+00:00

    Answer:

    a. u = 21. 50 m/s  

    Maximum height from the base of the cliff = 32 + 23.56 = 55.56  m

    Explanation:

    mass of rock = 0.26 kg

    Height = 32 m (the cliff cannot be 32 mm it should be 32 m)

    initial velocity (u) = ?

    final velocity(v) = 33 m/s

    Initial mechanical energy is equal to final mechanical energy. The object undergoes potential and kinetic energy.

    initial mechanical energy = final mechanical energy

    mgh + 1/2mu² = mgh + 1/2mv²

    mgh + 1/2mu² = 0 + 1/2mv²

    u² = v² -2gh

    u = √v² -2gh

    u = √1089 – 2 × 9.81 × 32

    u = √1089 – 627.84

    u = √461.16

    u = 21.4746362018

    u = 21. 50 m/s  

    b .The greatest maximum height

    1/2 mu² = mgh

    1/2u² = gh

    h = u²/2g

    h = (21.50)²/2 × 9.81

    h = 462.25
    /19.62

    h = 23.5601427115  m

    Maximum height from the base of the cliff = 32 + 23.56 = 55.56  m

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