Question

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

Answer 1

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.