How Can the Slope of an Incline Plane Affect the Motion of a Rolling Object?

[erok81]

Homework Statement

A snowball rolls off a roll that has a slope of 30* from the horizontal with a speed of 5.0 m/s. The edge of the roof is 10m above the ground. How far from the house (horizontally) does the snowball strike the ground?

Homework Equations

I'm using one of the kinematic equations - s_f=s_i+v_i t+1/2a(t)^2

The Attempt at a Solution

I know I need solve for both x and y planes. Solving for t on the y-axis gives me 3.016s to the ground.

Then plugging that in and using the same equation but solving for s_f...I get the wrong answer.

I know the angle of the roof the snowball falls from comes into play, but I don't know how. I thought it had something to do with the original position, but it doesn't. The only thing we've learned about inclined planes is their effect on gravity. Which doesn't apply to this problem.

Any pointers on how that roof slope comes into play?

 

[AEM]

Homework Statement

A snowball rolls off a roll that has a slope of 30* from the horizontal with a speed of 5.0 m/s. The edge of the roof is 10m above the ground. How far from the house (horizontally) does the snowball strike the ground?

Homework Equations

I'm using one of the kinematic equations - s_f=s_i+v_i t+1/2a(t)^2

The Attempt at a Solution

I know I need solve for both x and y planes. Solving for t on the y-axis gives me 3.016s to the ground.

Then plugging that in and using the same equation but solving for s_f...I get the wrong answer.

I know the angle of the roof the snowball falls from comes into play, but I don't know how. I thought it had something to do with the original position, but it doesn't. The only thing we've learned about inclined planes is their effect on gravity. Which doesn't apply to this problem.

Any pointers on how that roof slope comes into play?

Well, what you need to do is to use two of your kinematic equations. One for the horizontal direction and one for the vertical direction. You assume that the acceleration in the horizontal direction is zero (neglect air resistance). and take the vertical acceleration to be the acceleration of gravity. You will need the initial velocities in the horizontal and vertical directions. You get these from the slope of the roof and the velocity with which the ball leaves the roof. (Use a little trig to figure those out).

That should get you started.

 

[erok81]

Ooooh I think I get it. I was using 5.0 m/s as the initial velocity for both directions. But it isn't...at least I think it isn't.

So initial velocity is 2.5 downward and 4.33 horizontally.

Is that what you were referring to?

 

[erok81]

Ok...any more hints?

I tried it that way and am still wrong.

 

[AEM]

Ooooh I think I get it. I was using 5.0 m/s as the initial velocity for both directions. But it isn't...at least I think it isn't.

So initial velocity is 2.5 downward and 4.33 horizontally.

Is that what you were referring to?

That is what I was referring to. Your values for the initial velocity components are correct. I would place the origin of my coordinate system either at the point on the roof where the ball leaves the roof, or directly under it on the ground. Now write two equations:

X = X_0 + V_{0x} t

and

Y = Y_0 + V_{0y}t - \frac{1}{2} g t^2

Don't forget to put in the proper sign for your initial velocities.

You can eliminate the t variable and solve for the unknown that you want.