# Recent content by chocokat

1. ### Free-fall Acceleration Problem

Yes, just be sure you understand at what point in the ascent (or descent) the velocity represents. Once you have that, you should be able to use that information to find the answer you are looking for.
2. ### Free-fall Acceleration Problem

I'm going to jump in and try to help you along. You won't be able to do this in one easy step, so let's look at what you have. At the bottom of the window the flowerpot has some velocity v0. It goes up, above the window, at some top point has a final velocity of 0, then comes back down and...
3. ### Momentum HW problem help

You just have a sign problem. If Al moves in one direction at .897m/s, and Jo moves in the 'opposite' direction, the sign of Jo's velocity should be negative (-1.09m/s). Try that and it should work.
4. ### Momentum HW problem help

So, m1 + m2 = 167kg, rearrange the equation. m1 = ?
5. ### Projectile Motion Problem with thrown ball

KMJ - Snazzy is right, recalculate the velocity in the 'y' direction. 16.5(cos 36.44) does not equal 5.06 Snazzy - in projectile motion the 'x' direction velocity remains constant (well, for most of our calculations it does, the problems generally say to ignore wind resistance etc). The...
6. ### Impulse and Momentum Problem!

I think you've mixed up your sines and cosines. The 'x' components should be cosine values, and the 'y' components should be sine values. To start you off, based on the picture: Ax = - (500 cos 30) Ay = 500 sin 30
7. ### Uniform object acceleration

Yeah, you've got it now.
8. ### Uniform object acceleration

This is the equation you are using: What is the value of xi, and what is the value you used when you calculated it? (I'm sorry, I used x1 above when I should have used xi)
9. ### Uniform object acceleration

The first 16 should be x1, i.e. 3. Be careful of these little errors!
10. ### Football Throw

No, that's the answer I got doing it the other way, but I'm not familiar with that equation either.
11. ### Football Throw

No. You're basically starting over with t as an unknown. I did the problem myself this way. It's not too bad and really shouldn't take too long, I don't think as long as finding the height then the time the way you did. To get you started a bit, write down the variables you already know...
12. ### Football Throw

This way may be a little easier, and I got an answer that works. Just use the kinematic equation: \Delta s = v_i t + \frac{1}{2} a t^2 for each direction x and y. For each velocity, separate v_i into components. Then you have two equations with two unknowns, so you can then solve for...
13. ### Gravitational potential energy of skier

Assuming the skier started at height 0, then that answer looks right to me.
14. ### Gravitational potential energy of skier

Another note, although they are the same units, you are looking for gravitational potential energy, not work, so your formula isn't quite right. GPE = mgh So, you need to find 'h'. If the angle is 11.9˚ with the horizontal, how would you find the height?
15. ### Displacement vectors help

Not sure what you mean by this. The sin 90° = 1, so 3sin90 = 3*1 = 3 Be careful, the x value is 2.6, the y value is 4.5, and tan is y/x (not x/y as you have it). So, your angle isn't right. As to what to subtract it from, look at where it would be based on the signs of the x and y. Both are...