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17 February, 05:48

An object falls a distance h from rest. If it travels 0.460h in the last 1.00 s, find (a) the time and (b) the height of its fall.

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Answers (2)
  1. 17 February, 08:03
    0
    a) t = 3.771s

    b) h = 69.68 m

    Explanation:

    Given;

    total distance covered = h

    Using the second equation of motion;

    h = ut + 0.5at² ... 1

    We know u=0, since the object fall from rest.

    and also acceleration a = g (acceleration due to gravity)

    The equation 1 becomes;

    h = 0.5gt² ... 2

    Equation 2 is the overall equation of motion of the object.

    Where;

    h = height of fall

    t = total time of fall

    The motion have two phases.

    1) distance travelled in the first phase is

    h₁ = h - 0.46h = 0.54h

    Time taken for the first phase is;

    t₁ = t-1

    2) distance travelled in the second phase is

    h₂ = 0.46h

    Time taken for second phase is

    t₂ = 1.00s

    Writing the equation of motion of the first phase where the object starts from rest, we have;

    0.54h = 0.5g (t-1) ² ... 3

    Substituting equation 2 into 3, we have;

    0.54 (0.5gt²) = 0.5g (t-1) ²

    divide both sides by 0.5g

    0.54t² = (t-1) ²

    (t-1) ² - 0.54t² = 0

    Simplifying the equation above, we have;

    t² - 2t + 1 - 0.54t² = 0

    Then we have a quadratic equation;

    0.46t² - 2t + 1 = 0

    solving the quadratic equation, we have

    t = 3.771 or t = 0.5764

    Since t cannot be less than 1,

    Recall that t₁ = t-1, and t₁ cannot be negative. so t>1

    t = 3.771s

    b) Using equation 2

    h = 0.5gt²

    h = 0.5*9.8*3.771²

    h = 69.68 m
  2. 17 February, 08:49
    0
    A. Time, t = 4.35s

    B. Height of its fall, S = 92.72m

    Explanation:

    Vo = 0 m/s

    Vi = 0.46h m/s

    S = h m

    a = 9.81 m/s2

    To calculate the time taken, we need to get the value of the distance, h.

    Using the equations of motion,

    Vi^2 = Vo^2 + 2aS

    Where Vi = final velocity

    Vo = initial velocity

    a = acceleration due to gravity

    S = height of its fall

    (0.46h) ^2 = 0 + 2*9.81*h

    0.2116h^2 = 19.62h

    h = 19.62/0.1648

    = 92.722 m

    To calculate the time,

    S = Vo*t + (1/2) * a*t^2

    92.772 = 0 + (1/2) * 9.81*t^2

    t^2 = 185.44/9.81

    = 18.904

    t = sqrt (18.904)

    = 4.348 s
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