A stone with a weight of 5.29 N is launched vertically from ground level with an initial speed of 26.0 m/s, and the air drag on it is 0.262

Question

A stone with a weight of 5.29 N is launched vertically from ground level with an initial speed of 26.0 m/s, and the air drag on it is 0.262 N throughout the flight. What are (a) the maximum height reached by the stone and (b) its speed just before it hits the ground

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Doris 4 years 2021-08-25T10:05:06+00:00 1 Answers 14 views 0

Answers ( )

  1. bonexptip
    0
    2021-08-25T10:06:52+00:00
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    Answer:

    a) 19.4 m/s

    b) 19 m/s

    Explanation:

    a) In the given question,

    the potential energy at the initial point = Ui = 0

    the potential energy at the final point = Uf = mgh

    the kinetic energy at the initial point = Ki = 1/2 mv₀².

    the kinetic energy at the final point = Kf = 0

    work done by air= Ea= fh =  0.262 N

    Now, using the law of conservation of energy

    initial energy= final energy

    Ki +Ui = Kf + Uf +Ea

    1/2 mv₀² + 0 = 0 + mgh + fh

    1/2 mv₀² = mgh + fh

    h = v₀²/ 2g (1 +f/w)

    calculate m

    m= w/g = 5.29 /9.8

    = 0.54 kg

    h = 20 ²/ (2 x9.80) x (1 0.265/5.29)

    h = 19.4 m.

    b) 1/2 mv² + 2fh = 1/2 mv₀²

    Vg = 19 m/s

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