A particle is projected with a velocity of 29.4ms^-^1 . Find it’s maximum range on a horizontal plane through the point of projec

Question

A particle is projected with a velocity of 29.4ms^-^1 . Find it’s maximum range on a horizontal plane through the point of projection.
A.88.2m B.44.1m C.32.6m D.29.4m E.14.7m

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Lệ Thu 5 years 2021-07-23T22:56:47+00:00 2 Answers 41 views 0

Answers ( )

  1. Maris
    0
    2021-07-23T22:57:49+00:00
    Reply

    A.88.2m

    Answer:

    Solution given:

    initial velocity[u]=29.4m/s

    g=9.8m/s²

    maximum range=?

    now

    we have

    \theta=90°

    maximum range =\frac{29.4²*sin90}{9.8}=88.2m

  2. ngocdiep
    0
    2021-07-23T22:58:11+00:00
    Reply

    The initial velocity is,

    → u = 29.4 m/s

    General assumption,

    → g = 9.8m/s²

    → θ = 90°

    Then the maximum range is,

    → (29.4² × sin90)/9.8

    → 88.2 m

    Hence, option (A) is answer.

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