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    Physics Problems: Kinetics and Gravity

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    https://brainmass.com/physics/gravity/physics-problems-kinetics-gravity-35055

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    1.)
    Height of the building h = 16 m
    g = 9.8 m
    time of fall through the height of the building t = sqrt(2*h/g)
    => t = sqrt(2*16/9.8) = 1.80 sec

    At the time when 1st drop toucues the ground, the fifth one is leaving the roof. Therefore, there
    should be three drops in between 1st and 5th. Hence,
    t = 1.80 s = 4*T
    => T = interval between two successive drops = 1.80/4 = 0.45 s

    It means, at the time when 1st drop landing, the second drop has travelled by 3*T time, 3rd by 2*
    T, and 4th by T time.

    Hence, the position of 2nd drop from the ground
    h2 = 16 - (1/2)*g*(3*T) = 16 - 0.5*9.8*3*0.45 = 9.385 m --Answer

    3rd drop,
    h3 = 16 - 0.5*9.8*2*0.45 = 11.59 m --Answer

    4th drop,
    h4 = 16 - 0.5*9.8*0.45 = 13.795 m --Answer

    2.)
    Initital velocity u = vo (upward)
    Final velocity v = ?
    displacement s = 0 (coming back to the original point)
    acceleration a = g (downward)

    Because,
    v^2 = u^2 + 2*a*s
    => v^2 = vo^2 + 2*(-g)*0
    => v = - vo
    Here, -ve sign shows that the object will have doenward velocity.

    Magnitude of v = vo --Proved

    For rising,
    u = vo (upward)
    At the top most point, velocity v = 0
    time of rise t = t1 = ?
    acceleration a = g (downward)
    Because,
    v = u + a*t
    => 0 = vo - g*t1
    => t1 (time of rise) = vo/g

    Height of the rise h =?
    v^2 = u^2 + 2*a*s
    => 0 = vo^2 - 2*g*h
    => h = vo^2/2g

    Now, for fall,
    u = 0
    a = g (downward)
    displacement s = rise = h = vo^2/2g

    Hence ...

    Solution Summary

    The solution provides step-by-step workings for a number of physics-related calculations. The main concepts covered are kinetics and gravity.

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