Solving the Rock's Final Velocity Off a 33m Cliff

[AddversitY]

Homework Statement

A boy throws a rock off of a cliff that is 33m high. The rock weights 1 kilogram and it's inital velocity is 13.5m/s. It is thrown at an angle of 40° with the horizon.

Initial Velocity = 13.5m/s
mass = 1 kg
θ=40°
Δy = 33m
g = 9.81 m/s^2

Homework Equations

Potential Energy (Joules) = Ug = mgy
Kinetic Energy (Joules) = K = .5(m)(v^2)

The Attempt at a Solution

Calculate the velocity of the rock just before it hits the ground at bottom of cliff. No heat is generated--assume ideal conditions.

Potential Energy -> (Ug = mgy)
= (1kg)(9.81m/s^2)(33m)
= 323.4J

Kinetic Energy -> (K = .5(m)(v^2)
= .5(1kg)(13.5m/s^2)
= 91.125 J

This is where I get lost, I don't understand how to solve for the END velocity. I also didn't incorporate anything with the angle...I just feel lost and I would greatly appreciate assistance.

 

[cepheid]

Homework Statement

A boy throws a rock off of a cliff that is 33m high. The rock weights 1 kilogram and it's inital velocity is 13.5m/s. It is thrown at an angle of 40° with the horizon.

Initial Velocity = 13.5m/s
mass = 1 kg
θ=40°
Δy = 33m
g = 9.81 m/s^2

Homework Equations

Potential Energy (Joules) = Ug = mgy
Kinetic Energy (Joules) = K = .5(m)(v^2)

The Attempt at a Solution

Calculate the velocity of the rock just before it hits the ground at bottom of cliff. No heat is generated--assume ideal conditions.

Potential Energy -> (Ug = mgy)
= (1kg)(9.81m/s^2)(33m)
= 323.4J

Kinetic Energy -> (K = .5(m)(v^2)
= .5(1kg)(13.5m/s^2)
= 91.125 J

This is where I get lost, I don't understand how to solve for the END velocity. I also didn't incorporate anything with the angle...I just feel lost and I would greatly appreciate assistance.

You can examine the horizontal motions and vertical motions independently.

The horizontal velocity never changes, because there is no acceleration in the horizontal direction, because there is no net force in the horizontal direction. Therefore, the horizontal component of the velocity at the end of the fall is the same as it is at the beginning of the fall.

The vertical velocity changes, but you can easily use kinematics to solve for it. The downward force is constant, therefore, the acceleration is constant. What is the equation for velocity vs. time in the case of constant acceleration? Your vertical velocity will obey this equation.

Alternatively, you can use energy methods to solve for the final vertical velocity. Given the object's initial upward velocity, what maximum height does it reach (use conservation of energy)? Since it then falls to the ground from that height, what velocity will it have at the end of the fall? Again, use conservation of energy.

Edit: the second method is definitely preferable, because you don't have to worry about the time interval involved.

 

[AddversitY]

You can examine the horizontal motions and vertical motions independently.

The horizontal velocity never changes, because there is no acceleration in the horizontal direction, because there is no net force in the horizontal direction. Therefore, the horizontal component of the velocity at the end of the fall is the same as it is at the beginning of the fall.

The vertical velocity changes, but you can easily use kinematics to solve for it. The downward force is constant, therefore, the acceleration is constant. What is the equation for velocity vs. time in the case of constant acceleration? Your vertical velocity will obey this equation.

Alternatively, you can use energy methods to solve for the final vertical velocity. Given the object's initial upward velocity, what maximum height does it reach (use conservation of energy)? Since it then falls to the ground from that height, what velocity will it have at the end of the fall? Again, use conservation of energy.

Edit: the second method is definitely preferable, because you don't have to worry about the time interval involved.

Thank you for the response, it is greatly appreciated! However, I did not specify that I have the knowledge to solve this problem using Kinematic Equations. I apologize!

Here is new attempt to solve the problem using Conservation of Energy.

Energy (E) = Kinetic Energy (K) + Potential Energy (U)

The initial Kinetic Energy (Ki) is as follows:
Ki = .5(m)(v^2)
= .5(1kg)(13.5m/s)^2
= 91.125 Joules

The initial Potential Energy (Ui) is as follows:
Ui = mgy
= (1kg)(9.8m/s^2)(33m)
= 323.4 Joules

Thus, the initial Mechanical Energy is Ui + Ki.
Mechanical Energy (ME) = 414.525 J

The Law of Conservation of Energy yields initial Energy (Ei) is equal to final Energy (Ef) in a system.

So, if there is 414.525 Joules of energy being produced at the initial, the same must stand for the final.

At Uf Final Potential Energy, height = 0.

Uf = mgy
Uf = (1kg)(9,81m/s^2)(0m)
= 0

All the energy must therefore be in the form of Kinetic Energy.
Substituting that in and solving for final velocity..

414.525 J = .5(1kg)(v)^2
v^2 = 829.05 (Dividing through by .5)
v = +- 28.8 m/s (Taking square root)

Would this value be negative or positive? Also, was the process correct? Why do I not have to take into consideration the angle if so?

I appreciate the help.

 

[cepheid]

Thank you for the response, it is greatly appreciated! However, I did not specify that I have the knowledge to solve this problem using Kinematic Equations. I apologize!

Here is new attempt to solve the problem using Conservation of Energy.

Energy (E) = Kinetic Energy (K) + Potential Energy (U)

The initial Kinetic Energy (Ki) is as follows:
Ki = .5(m)(v^2)
= .5(1kg)(13.5m/s)^2
= 91.125 Joules

You should just consider the kinetic energy due to vertical motion. There are no horizontal forces, and hence none of the work that is done changes the horizontal speed. Only the vertical speed changes.

Use the angle to compute the vertical component of the initial velocity. Then plug that into the kinetic energy equation.

The result for the final vertical speed is no different than it would be if you had thrown the ball straight up with this initial vertical speed.

 

[AddversitY]

You should just consider the kinetic energy due to vertical motion. There are no horizontal forces, and hence none of the work that is done changes the horizontal speed. Only the vertical speed changes.

Use the angle to compute the vertical component of the initial velocity. Then plug that into the kinetic energy equation.

The result for the final vertical speed is no different than it would be if you had thrown the ball straight up with this initial vertical speed.

Here is new NEW attempt to solve the problem using Conservation of Energy.

The initial Velocity in the y direction is:
Vy = 13.5m/s*sin(40°)
Vy = 8.7 m/s


Energy (E) = Kinetic Energy (K) + Potential Energy (U)

The initial Kinetic Energy (Ki) is as follows:
Ki = .5(m)(v^2)
= .5(1kg)(8.7m/s)^2
= 37.845 J

The initial Potential Energy (Ui) is as follows:
Ui = mgy
= (1kg)(9.8m/s^2)(33m)
= 323.4 Joules

Thus, the initial Mechanical Energy is Ui + Ki.
Mechanical Energy (ME) = 361.245 J

The Law of Conservation of Energy yields initial Energy (Ei) is equal to final Energy (Ef) in a system.

So, if there is 361.245 Joules of energy being produced at the initial, the same must stand for the final.

At Uf Final Potential Energy, height = 0.

Uf = mgy
Uf = (1kg)(9,81m/s^2)(0m)
= 0

All the energy must therefore be in the form of Kinetic Energy.
Substituting that in and solving for final velocity..

361.245 J = .5(1kg)(v)^2
v^2 = 722.49 (Dividing through by .5)
v = +- 26.88m/s (Taking square root)

Like this?

 

[cepheid]

Your method seems fine. The final answer you get is the final *vertical* velocity on impact. You need to combine this (vectorially of course) with the horizontal velocity and then compute the magnitude of the resultant in order to get the total velocity upon impact.

Regarding the sign of the answer: it's up to you which sign convention you pick. You can either choose "downward" to be the negative vertical direction or the positive vertical direction. Just choose one and use it consistently.

If you know how to solve the problem using kinematics, then why not just do that in order to double check your answer?

 

[BeBattey]

I think what hung you up is what the mgy is symbolizing.

In the distance y you have from the ground, you have mgy joules of energy that the gravitational force will put into your object before it smacks into the ground. Where y=0 is completely, utterly up to you. If you set the ground equal to y=17, and the boy on top of the cliff to be y=50, you would come up with the same answer, but in the end you will have KE - GPE, it's just convenient to set the ground to y=0 so your end equation would be (1/2)mv^2 - mg(0) = mgy, or just (1/2)mv^2 = mgy as opposed to (1/2)mv^2 - 17mg = 50mg

Hopefully that helped!

 

[AddversitY]

Okay, on another note: If there rock were to hit the ground at a velocity of 25m/s, what would be the heat loss.

To attack this problem, I'd find initial kinetic and initial potential and add them to get initial mechanical energy.

Initial:
K = .5(m)(v)^2
= .5(1kg)(13.5m/s)^2
= 91.125 J

U = mgy
= (1kg)(9.81m/s^2)(33m)
= 323.4J

Initial Mechanical Energy is *(414.525 J)*.

Final:
K = .5(m)(v)^2
= .5(1kg)(25m/s)^2
= 312.5 J

U = mgy
= (1kg)(9.8m/s^2)(0m)
= 0 J

Final Mechanical Energy is 312.5 J.

Since all energy is conserved:

414.525 J - 312.5 J = 102.025 J

102.025 J lost to heat.

Correct?^

 

[cepheid]

Okay, on another note: If there rock were to hit the ground at a velocity of 25m/s, what would be the heat loss.

To attack this problem, I'd find initial kinetic and initial potential and add them to get initial mechanical energy.

Initial:
K = .5(m)(v)^2
= .5(1kg)(13.5m/s)^2
= 91.125 J

U = mgy
= (1kg)(9.81m/s^2)(33m)
= 323.4J

Initial Mechanical Energy is *(414.525 J)*.

Final:
K = .5(m)(v)^2
= .5(1kg)(25m/s)^2
= 312.5 J

U = mgy
= (1kg)(9.8m/s^2)(0m)
= 0 J

Final Mechanical Energy is 312.5 J.

Since all energy is conserved:

414.525 J - 312.5 J = 102.025 J

102.025 J lost to heat.

Correct?^

Seems okay to me.

 

[AddversitY]

Your method seems fine. The final answer you get is the final *vertical* velocity on impact. You need to combine this (vectorially of course) with the horizontal velocity and then compute the magnitude of the resultant in order to get the total velocity upon impact.

Regarding the sign of the answer: it's up to you which sign convention you pick. You can either choose "downward" to be the negative vertical direction or the positive vertical direction. Just choose one and use it consistently.

If you know how to solve the problem using kinematics, then why not just do that in order to double check your answer?

If 26.88m/s is the vertical vector and I find initial horizontal velocity velocity (10.34m/s) and use Pythagorean theorem I come up with the same answer I got before...28.8m/s. Does this seem to be the correct velocity of the rock right before it hits the ground?

 

[PhanthomJay]

If 26.88m/s is the vertical vector and I find initial horizontal velocity velocity (10.34m/s) and use Pythagorean theorem I come up with the same answer I got before...28.8m/s. Does this seem to be the correct velocity of the rock right before it hits the ground?

When you solved the problem using your method in post #3, you correctly determined the magnitude of the final velocity (its speed), but you said nothing about its direction. When solving the problem using Cepheid's approach, you have correctly determined the the final vertical velocity component and the final horizontal velocity component, and the magnitude of the final velocity using pythagoras, but you still have not directly specified the final velocity direction, using trig.

 

[The Anonymous]

The beauty of the work-energy theorem (assuming only conservative forces act on the system) is that energy is a scalar, so you don't need any vector components in order to solve the problem.
Initial kinetic + initial potential = final kinetic + final potential
final kinetic = Initial kinetic + initial potential - final potential
Final kinetic = 0.5*m*v^2

Solve for v.

Edit: This v is a speed, not a velocity - as has been mentioned. Are you tasked with finding speed or velocity just prior to impact?
If speed, then you are done. If velocity, then you could use trig, realizing that the horizontal component is assumed to be invariant during flight.cheers