A ballast is dropped from a stationary hot-air balloon that is at an altitude of 576 ft. Find (a) an expression for the altitude of the ball

Question

A ballast is dropped from a stationary hot-air balloon that is at an altitude of 576 ft. Find (a) an expression for the altitude of the ballast after t seconds, (b) the time when it strikes the ground, and (c) its velocity when it strikes the ground. (Disregard air resistance and take ft/sec2.)

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Minh Khuê 4 years 2021-08-15T11:08:10+00:00 1 Answers 101 views 0

Answers ( )

  1. ThanhThu
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    2021-08-15T11:09:23+00:00
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    Answer:

    a) S = \frac{1}{2}gt^2\\

    b) 6secs

    c) 192ft

    Explanation:

    If a ball dropped from a stationary hot-air balloon that is at an altitude of 576 ft, an expression for the altitude of the ballast after t seconds can be expressed using the equation of motion;

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

    S is the altitude of the ballest

    u is the initial velocity

    a is the acceleration of the body

    t is the time taken to strike the ground

    Since the body is dropped from a stationary air balloon, the initial velocity u will be zero i.e u = 0m/s

    Also, since the ballast is dropped from a stationary hot-air balloon, the body is under the influence of gravity, the acceleration will become acceleration due to gravity i.e a = +g

    Substituting this values into the equation of the motion;

    S = 0 + \frac{1}{2}gt^2\\ S = \frac{1}{2}gt^2\\

    a) An expression for the altitude of the ballast after t seconds is therefore

    S = \frac{1}{2}gt^2\\

    b) Given S = 576ft and g = 32ft/s², substituting this into the formula in (a);

    576 = \frac{1}{2}(32)t^2\\\\\\576*2 = 32t^2\\1152 = 32t^2\\t^2 = \frac{1152}{32} \\t^2 = 36\\t = \sqrt{36}\\ t = 6.0secs

    This means that the ballast strikes the ground after 6secs

    c) To get the velocity when it strikes the ground, we will use the equation of motion v = u + gt.

    v = 0 + 32(6)

    v = 192ft

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