A rock drops from the top of a 10.5 m building. What is the velocity when he hits the ground?what is the Plot the position, velocity and acc

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A rock drops from the top of a 10.5 m building. What is the velocity when he hits the ground?what is the Plot the position, velocity and acceleration vs. time

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Mộc Miên 6 months 2021-07-13T04:39:51+00:00 1 Answers 4 views 0

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    2021-07-13T04:40:54+00:00

    Answer:

    When the rock is on top of the building, it does not move, so it only has potential energy.

    The potential energy can be written as:

    U = m*g*h

    where m is the mass, g is the gravitational acceleration, h is the height.

    Now, as the rock starts to fall down, the potential energy transforms into kinetic energy.

    K = (m/2)*v^2

    Where v is the velocity.

    When the rock hits the ground, all the potential energy has ben converted into kinetic energy, then:

    U = K

    m*g*h = (m/2)*v^2

    Here we can isolate v:

    v = √(2*g*h)

    and g = 9.8m/s^2

          h = 10.5m

    v = √(2*10.5m*9.8m/s^2) = 14.34m/s

    Now the second question:

    “what is the Plot the position, velocity and acceleration vs. time”

    I suppose that you need to select the correct plot for each thing, the images are not given, so let’s analyze how each plot is:

    The motion equations are:

    Acceleration:

    Here we have only the gravitational acceleration, so we can write:

    a(t) = -g

    This is a constant, the graph will be a horizontal line at y = -g.

    Velocity:

    We integrate the acceleration over time, the constant of integration is the initial velocity, that in this case is zero.

    v(t) = -g*t

    This is a linear equation with slope equal to -g, and y-intercept equal to zero.

    Position.

    We integrate again over time, this time the constant of integration will be the initial height of the rock = 10.5m

    The equation is:

    p(t) = -(g/2)*t^2 + 10.5m

    This is a quadratic equation with a negative leading coefficient, so the arms go downwards.

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