# Kinematic Energy Problem

1. Dec 14, 2011

### landlord

1. The problem statement, all variables and given/known data

Santa has an unfortunate roof mishap. When he steps out of his sleigh, he realizes that there is a coating of ice on this particular roof—making it a completely frictionless surface. Starting with an initial velocity of zero, he slides down the roof (a distance of 2.50 m along the rooftop) before he becomes a projectile. He lands in a soft pile of snow that is 15.0 m lower than the edge of the roof. How far from the house is the pile of snow in which Santa lands (what is the range)? During his slide on the roof, his acceleration is 1.9 m/s2 and the angle that the roof top makes with the horizontal is 30.0°.

Vo = 0m/s
Vf = ?
a = 1.9 m/s2

2. Relevant equations

V=Vo + at
Δx=Vot + 1/2at^2

3. The attempt at a solution
My professor said that this was an energy conservation problem using kinematic equations, but I don't see that.

V=Vo+at
t = V-Vo/a (getting rid of the t variable)

Δx=Vo(V-Vo/a) + 1/2a(V-Vo/a)^2

Am I on the right track? Do I need to break these up into Y and X components?

2. Dec 14, 2011

### netgypsy

You have a kinematics equation that is actually derived from an energy equation
vfinal ^2 = v initial ^2 + 2aL where L is the displacement traveled. If you did your two equation above correctly this is probably what you will get.

Remember he started at rest, you know the length he slid, the acceleration, so that enables you to find the final velocity of the slide which is the initial velocity of the launch. Since you know the roof angle you also know the angle of the launch velocity.

From there there are a number of ways to solve for how far from the house he landed but do remember that his launch velocity has both x and y components and his landing point is dependent on the time before he hits and the x component of his velocity. There is a range equation but I never used it but you can if you use it right.

3. Dec 14, 2011

### Delphi51

Caution: you can use conservation of energy to find that speed at the edge of the roof but that way you'll get a different answer compared to using the a = 1.9 due to an error in the question. For a 30 degree roof, the acceleration along the roof should be 9.8*sin(30) = 4.9 rather than 1.9.

4. Dec 14, 2011

### LawrenceC

"My professor said that this was an energy conservation problem using kinematic equations, but I don't see that."

Problem can be done either way. With the energy approach, you consider Santa's initial and final kinetic energies in the vertical direction (his horizontal velocity is constant once he leaves the roof) as well as his initial and final potential energies. Having done the above, you can compute his final vertical velocity when he hits the ground. Because gravity is constant, you can average the velocities to obtain a mean vertical speed. From that you can easily complete the problem.

5. Dec 14, 2011

### netgypsy

I wondered about that myself cause no way a = 1.9m/sec^2 on a 30 degree roof with no friction. I wondered why a was given in the first place???

6. Dec 14, 2011

### landlord

Thanks for all the feedback, it sounds like there are a few ways to do this problem. I'm still kind of confused with the question because I have never done like this before. Do any of you guys mind laying out how to do this problem? This is a review for my final exam, so that could be why the 1.9 is incorrect.

7. Dec 14, 2011

### Delphi51

Recommend you do not use the 1.9. Do the energy thing as suggested by the prof:
Potential energy at the top = kinetic energy at the bottom of the roof

8. Dec 14, 2011

### landlord

mghf + .5mv^2f = mgho + (.5mv^2o == 0 because initial velocity = 0)

mghf + .5mv^2f = mgho

is this the correct way to set up the equation?

9. Dec 14, 2011

### netgypsy

that will work. Just be sure to get the vertical height down the roof by using the angle of the roof and the distance sliding down the roof and the correct trig function. This will be delta h or the vertical height Santa actually falls as he slides down the roof.

10. Dec 14, 2011

### landlord

this is what i have done so far, can u help me input where the delta h would be in the equation

.5mvf ^2 + mghf = mgho
.5mvf^2 = mgho - mghf
mvf^2 = 2mgho - mghf
vf^2 = (2mgho - mghf)/m
vf = sqrt[(2mgho-mghf)/m]

i hope this is right :/

11. Dec 14, 2011

### netgypsy

You lost a 2 when you divided through by a .5.
You'll get mvf^2 = (2mgho - TWOmghf)/m

factor out the 2mgh and you'll get
mvf^2 = 2mgh(ho - hf)/m and ho - hf is the change in altitude (delta h ) from the peak of the roof to the eaves which you can find using the slide length and the roof angle This is the vertical height Santa drops.