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18 February, 07:12

1. a bag is dropped from a hovering helicopter. the bag has fallen for 2 s. what is the ball's velocity at the instant its hitting the ground? what was the height of the helicopter? if the bag was dropped from a helicopter descending at the speed of 2 m/s what is the time of the bag fall and its velocity near the ground?

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  1. 18 February, 07:18
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    1. The bag's velocity immediately before hitting the ground.

    Recall this kinematics equation:

    Vf = Vi + aΔt

    Vf is the final velocity, Vi is the initial velocity, a is the acceleration, and Δt is the time elapsed.

    Given values:

    Vi = 0m/s (you assume this because the bag is dropped, so it falls starting from rest)

    a is 9.81m/s² (this is the near-constant acceleration of objects near the surface of the earth)

    Δt = 2s

    Plug in the values and solve for Vf:

    Vf = 0 + 9.81*2

    Vf = 19.62m/s

    2. The height of the helicopter.

    Recall this other kinematics equation:

    d = ViΔt + 0.5aΔt²

    d is the distance traveled by the object, Vi is the initial velocity, a is the acceleration, and Δt is the time elapsed.

    Given values:

    Vi = 0m/s (bag is dropped starting from rest)

    a = 9.81m/s² (acceleration due to gravity of the earth)

    Δt = 2s

    Plug in the values and solve for d:

    d = 0*2 + 0.5*9.81*2²

    d = 19.62m

    3. Time of the bag's fall and its velocity immediately before hitting the ground ... if it started falling at 2m/s

    Reuse the equation from question 2:

    d = ViΔt + 0.5aΔt²

    Given values:

    d = 19.6m (height of the helicopter obtained from question 2)

    Vi = 2m/s

    a = 9.81m/s² (acceleration due to earth's gravity)

    Plug in the values and solve for Δt:

    19.6 = 2Δt + 0.5*9.81Δt²

    4.91Δt² + 2Δt - 19.6 = 0

    Use the quadratic formula to get values of Δt (a quick Google search will give you the formula and how to use it to solve for unknown values):

    Δt = 1.8s, Δt = - 2.2s

    The formula gives us 2 possible answers for Δt but within the situation of our problem, only the positive value makes sense. Reject the negative value.

    Δt = 1.8s

    Now we can use this new value of Δt to get the velocity before hitting the ground:

    Vf = Vi + aΔt

    Given values:

    Vi = 2m/s

    a = 9.81m/s²

    Δt = 1.8s (result from previous question)

    Plug in the values and solve for Vf:

    Vf = 2 + 9.81*1.8

    Vf = 19.66m/s
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