# Derive an expression for the angle φ (Kinematics Problem)

1. Sep 17, 2016

### JohnTravolski

• Moved from a rechnical forum, so homework template missing
I'm really lost on this one.

I'm really not sure how to get started on this. I started it out as a force problem and solved for ax = g*cos(theta). I then integrated that (treating it like a constant) from zero to D to find my initial velocity at the time the block reaches the edge of the cliff. I then used the position function:
y=y0+v0yt+1/2ayt^2

to determine the time it takes to hit the ground, but I don't understand how any of this will help me find the angle φ. I'm really just lost. Can somebody please work this problem so I won't be so confused?

2. Sep 17, 2016

### drvrm

at least you should work out the first part and from geometry find the angle of projection from the height H.
that is take the projectile motion and the sliding motion in two frames;as per Hint.

3. Sep 17, 2016

### vela

Staff Emeritus
Why did you find the time?

There is a limited number of quantities in these kinematic problems. Knowledge of which ones would let you determine the angle $\phi$? That's what you need to figure out first, and then solve for those quantities using the kinematic equations.

4. Sep 18, 2016

### JohnTravolski

Well, before I get that far, do I have to first somehow calculate the initial velocity at the moment the box leaves the incline? If so, how do I go about that?

5. Sep 18, 2016

### J Hann

What "vertical" distance does the block travel while sliding down the incline?
What does conservation of energy tell you about the speed of the block as it leaves the incline?

6. Sep 18, 2016

### JohnTravolski

The vertical distance it travels is D*Cos(theta). The horizontal distance it travels is D*Sin(theta).

I'm still unsure of how to get velocity out of that since I'm not provided with any time interval.

7. Sep 18, 2016

### J Hann

What does conservation of energy tell you about potential energy and kinetic energy?

8. Sep 18, 2016

### JohnTravolski

None of it is lost. So that means that the velocity is related to the distances it travels.

9. Sep 18, 2016

### J Hann

Kinetic energy and hence velocity is related to the potential energy lost.
What is the change in potential energy of the block?

10. Sep 18, 2016

### vela

Staff Emeritus
Didn't you already do that? It sounds like it based on your initial post.

Anyway, the reason I asked you what you need to know to calculate the final answer is because knowing that guides you into figuring out what intermediate results you may need. For example, if all you needed to know was the speed of the object when it lands, finding the velocity as it leaves the cliff is a waste of time. So it's not really about "get[ting} that far." It's where you should be starting from.

11. Sep 18, 2016

### JohnTravolski

I thought I did calculate it but I have no idea if what I did was the correct method or not. I'm basically guessing my whole way through this problem. If what I did is correct, I'm still unsure of how to proceed.

12. Sep 18, 2016

### vela

Staff Emeritus
That's why you need to figure out what you need to know to calculate $\phi$. Knowing that will suggest to you how to proceed.

13. Sep 18, 2016

### JohnTravolski

The x and the y components of velocity at the time the box hits the ground.

14. Sep 18, 2016

### JohnTravolski

15. Sep 18, 2016

### vela

Staff Emeritus
Right.

You made a mistake in the beginning when you wrote that $v = \int a_x$. It's important to specify which variable you're integrating with respect to. It's correct to say that $v = \int a\,dt$, but you're calculating $\int a\,dx$. When you integrate with respect to position, you end up with the equation $v^2 = v_0^2 + 2a(x-x_0)$ for constant acceleration. (So you don't really need to integrate here. If you have constant acceleration, you can use that equation without re-deriving it each time.)

Your strategy for finding $v_y$ and $v_x$ at the end looks fine. You just need to fix the intermediate result you got for $v_0$, the speed at which the block leaves the incline. If you check the units in your current result, you'll see they don't work out, i.e., gD cos theta doesn't have units of velocity. If you're unsure of your work, checking units is a quick way to see if you possibly made a mistake.

16. Sep 18, 2016

### JohnTravolski

Okay, I think that I finally understand. I reworked the problem and this time my answer has the correct units (which all cancel out). Here's my new work:
https://i.imgsafe.org/f019ae72df.jpg
I'm pretty sure this is the correct answer. I don't know if it can be simplified or not.

17. Sep 18, 2016

### vela

Staff Emeritus
You're mixing up the x and y coordinates in the first part of the problem with the x and y coordinates of the second part. Let's use primes, e.g. x' and y', for the second half.

The speed you found $v_x = \sqrt{2gD\cos\theta}$ should be the hypotenuse of the triangle you drew halfway down the page, not the x'-component.

You should be able simplify the final expression. Because the acceleration of gravity appears in every term, it'll cancel. Same with the factor of 2.

18. Sep 18, 2016

Thank you.