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.
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...
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.
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...
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
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)
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...
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...
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?
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...
There are a couple of things going on here, one is I'm not quite sure how you're getting the answers you're getting.
First let's simplify things a bit. Vector B is vertical and on the y axis, therefore there is no x component, so you should just use 0 when referring to the x component of B...
Yes, your angle would just be 90 degrees, or the positive y axis.
For the second problem, I think you're just working too fast.
For the i's (-0-8-8) = -16
For the j's (-3--1+5) = 3
So the magnitude is \sqrt{(-16)^2 + 3^2}
which equals 16.28
I'm going to let you figure out the...
I did a quick check of your first answer, but it's wrong. You need to recheck your math.
For the i's (0+8-8) = 0
For the j's (3-1+5) = 7
So the magnitude of A+B+C = 7
Now, since the i's represent the x value, and the j's represent the y value, what do you know about the angle that has an x...
I can try to help. First, you should convert the distance in km to s, i.e. 384,000 km = ?s.
I used c = 3 * 10^8 and got d = 1.28s.
Now, I know that the aliens travelling in the ufo will have the shortest time. I used the spacetime equation
\Delta s^2 = \Delta t^2 - \Delta d^2
in this...
Yes, you're exactly on the right track. Plug it into the equation:
\Delta x = v_0 t + \frac{1}{2} a t^2
where a = 0 so that whole piece goes away, and solve for v and you're all set.
BTW, I got t = 4.04s which gets you much closer to the actual answer than just t = 4s.
I think you're making the problem more difficult than it is.
Based on the wording and picture of the problem, how much of initial velocity is horizontal, and how much is vertical? (Hint - you don't need trig functions).
For both of these problems you can use the equation:
{v_f}^2 = {v_0}^2 + 2as
Where v = final velocity
v_0 = initial velocity
s = change of distance
This is one of the standard constant acceleration formulas. But it has a caveat - you cannot use it if the direction changes (which it doesn't...
Yes.
Using the 60 degree angle, you can use the (what I'll call) traditional sin for the 'y' length and cos for the 'x' length. Now your calculations should work.
You can also use the 30 degree angle. But you'd have to use sin for the 'x' length, and cos for the 'y' length. It's always...
You have to be sure to read the problem correctly:
The vector makes an angle 30 degrees clockwise from the +y axis. If you want to use the +x axis as your guide, and you drop a perpendicular line from the vector, what is the angle from the +x axis?
You have the (almost) correct answer, but I would suggest you change your process a bit.
You're looking for the acceleration of 'you' so that the block does not move. So, you're main goal is:
\Sigma F = ma = 0
We can use m_1 for the person, and m_2 for the block. And we know
m_2 =...
For #3 I think you can use the equation:
v_2 = v_1 + at
where:
v_2 = 0 m/s (at max height velocity is 0 as antenna guy notes)
v_1 = 4.706 m/s (I'm assuming the calculation for this is correct)
a = -9.8 m/s^2
Then just solve for 't'.