A stone is thrown vertically upward with an initial velocity of 40m/s. Taking g = 10 m/s^2 find the maximum height reach by the stone and wh

Question

A stone is thrown vertically upward with an initial velocity of 40m/s. Taking g = 10 m/s^2 find the maximum height reach by the stone and what is the net displacement and distance covered by the stone.​

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Cherry 4 years 2021-07-16T23:40:06+00:00 2 Answers 21 views 0

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  1. Nem
    0
    2021-07-16T23:41:36+00:00
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    Answer:

    The maximum height is 80 m and the total distance covered by the stone is 160 m and the displacement is zero.

    Explanation:

    Final velocity v = 0

    Initial velocity u = 40m/s

    We know that,

    Using equation of motion

    v^{2} =u^{2} +2gh

    0-40^{2} =2 × 10 × h

    The maximum height is:

    h=80 m

    The  stone will reach at the top and will come down

    Therefore, the total distance will be:

    s=h_{1} +h_{2}

    s=80m-80m=160m

    The net displacement is:

    D=h_{1} -h_{2}

    D=80m-80m=0

    Hence, The maximum height is 80 m and the total distance covered by the stone is 160 m and the displacement is zero.

    hope this helps…..

  2. huyenthanh
    0
    2021-07-16T23:41:51+00:00
    Reply

    Explanation:

    u=40

    v=?

    h=?

    v²-u²=2gs

    0²-40²=2×10×s

    160=20s

    s=160/20

    =80m/s

    total distance= upward distance ×downward distance

    =80+80

    =160m

    total displacement=0 because u and v is the same.

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