A cannon fires a shell straight upward; 1.8 s after it is launched, the shell is moving upward with a speed of 17 m/s. Assuming air resistan

Question

A cannon fires a shell straight upward; 1.8 s after it is launched, the shell is moving upward with a speed of 17 m/s. Assuming air resistance is negligible, find the speed (magnitude of velocity) in meters per second of the shell at launch and 5.5 s after the launch

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Khánh Gia 5 years 2021-07-18T21:29:23+00:00 1 Answers 99 views 0

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  1. niczorrrr
    0
    2021-07-18T21:31:04+00:00
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    To solve this problem we will apply the concepts of acceleration according to the laws of kinetics. Acceleration can be defined as the change in velocity per unit time, that is,

    a = \frac{v-u}{t}

    Here,

    v = Final velocity

    u = Initial velocity

    t = Time

    Rearranging to find the initial velocity,

    u = v-at

    Now the acceleration is equal to the gravity, then,

    u = v+gt

    u = 17+(9.8)(1.8)

    u = 34.64m/s

    The velocity of the shell at 5.5s after the launch is,

    v = u+at

    v = u-gt

    v = 34.64-(9.8)(5.5)

    v = -19.26m/s

    Therefore the magnitude of the velocity of the shell at 5.5s is -19.26m/s

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