4. A rock is thrown from the edge of the top of a 100 m tall building at some unknown angle above the horizontal. The rock strikes the groun

Question

4. A rock is thrown from the edge of the top of a 100 m tall building at some unknown angle above the horizontal. The rock strikes the ground at a horizontal distance of 160 m from the base of the building 5.0 s after being thrown. Determine the speed with which the rock was thrown.

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niczorrrr 4 years 2021-08-27T06:59:14+00:00 2 Answers 119 views 0

Answers ( )

  1. Maris
    0
    2021-08-27T07:00:27+00:00
    Reply

    Answer:

    55.42 m/s

    Explanation:

    Along the horizontal direction, the rock travels at constant speed: this means that its horizontal velocity is constant, and it is given by

    u_x = d/t

    Where

    d = 160 m is the distance covered

    t = 5.0 s is the time taken

    Substituting, we get

    u_x =160/5 = 32 m/s.

    Along the vertical direction, the rock is in free-fall – so its motion is a uniform accelerated motion with constant acceleration g = -9.8 m/s^2 (downward). Therefore, the vertical distance covered is given by the

    S=u_yt+\frac{1}{2}at^2


    where


    S = -100 m is the vertical displacement


    u_y is the initial vertical velocity


    Replacing t = 5.0 s and solving the equation for u_y, we find

    -100 = u_y(5) + (-9.81)(5)^2/2


    u_y = 45.25 m/s

    Therefore, the speed with which the rock was thrown u

    u= \sqrt{u_x^2+u_y^2} \\=\sqrt{32^2+45.25^2}\\ = 55.42 m/s

  2. ngocdiep
    0
    2021-08-27T07:00:49+00:00
    Reply

    Answer:

    Explanation:

    Let the velocity of projectile be v and angle of throw be θ.

    The projectile takes 5 s to touch the ground during which period it falls vertically by 100 m

    considering its vertical displacement

    h = – ut +1/2 g t²

    100 = – vsinθ x 5 + .5 x 9.8 x 5²

    5vsinθ =  222.5

    vsinθ = 44.5

    It covers 160 horizontally in 5 s

    vcosθ x 5 = 160

    v cosθ = 32

    squaring and adding

    v²sin²θ +v² cos²θ = 44.4² + 32²

    v² = 1971.36 + 1024

    v = 54.73 m /s

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