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

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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