A.P Physics Help!!! (urgggg)

<p>Anyone willing to help me with some A.P physics problems dealing with kinematics? (These questions have all been on real A.P Physics B tests in the past, and I am having some difficulty with them. I would be really grateful if anyone is willing to help me solve these problems.
Either pm me or just reply to this thread and I will post them.</p>

<p>Post it up and I'll see what I can do.</p>

<p>I'm not in an AP class but I might be able to help.</p>

<p>thanks iljets... and sorry I didnt post it sooner....</p>

<p>A ball of mass 0.5 kilogram, initially at rest, is kicked directly toward a fence from a point 32 meters away, as shown above. The velocity of the ball as it leaves the kicker's foot is 20 meters per second at an angle of 37° above the horizontal. The top of the fence is 2.5 meters high. The kicker's foot is in contact with the ball for 0.05 second. The ball hits nothing while in flight and air resistance is negligible.</p>

<p>a. Determine the magnitude of the average net force exerted on the ball during the kick.
b. Determine the time it takes for the ball to reach the plane of the fence.
c. Will the ball hit the fence? If so, how far below the top of the fence will it hit? If not, how far above the top of the fence will it pass?</p>

<p>a. They're looking for net force? Okay. First, resolve your vectors.
1. Find the horizontal and vertical components of your velocity.
2. Find the separate acceleration of those components using their velocity component and the time given to you : 0.05s (acceleration is change in velocity divided by change in time)
3. Resolve the forces vertically and horizontally (F = ma but for vertical there is also gravity, so be careful)
4. Add the forces using vectors.</p>

<p>b. This one is quite straightforward. I will just say to use your kinematics equations..</p>

<p>c. This one is also using your kinematics equations.</p>

<p>From what I can tell, a is probably the one that needed the most thinking. Parts b and c you should be able to get after toying around with the equations for a while.</p>

<p>Let me know if you need more help. :)</p>

<p>Let's see... where to start....</p>

<p>I'll start with Part A. (sounds logical)</p>

<p>A ) You know that Force = mass * acceleration, and acceleration is the velocity divided by the time taken to achieve that velocity, or 20/0.05, or (breaks out calculator) 400 m/s^2. The mass of the ball is 0.5, so 0.5 x 400 = 200 Newtons.
B) You must break the 20 m/s into components, so 20*sin(37) = 12.04m/s, which is the y component of the velocity. The x component of the velocity is 20 * cos(37) = 15.97 m/s. Since the delta x = vot + 1/2 at^2, and you know that there will be no acceleration in the x direction (since your foot isn't continually kicking the ball), delta x = vot; dividing both sides by your initial velocity gives you: t = x/vo; t = 2s.</p>

<p>C) For this we will need the y component, so refer back to (b) for the y compenent of the velocity, which at the start is 12.04 m/s. We know that delta y = voyt + 1/2 ay t^2, and a = -9.81 m/s^2 (gravity).</p>

<p>We plug in: delta y = 12.04*2 + 1/2 * (-9.81) (2^2), which gives you: 4.46 meters. Since we know the fence is 2.5 meters high, 4.46 - 2.5 gives you 1.96 meters above the fence.</p>

<p>If any of this makes no sense and/or the answers are wrong...respond of course.</p>

<p>Edit: I see I am too slow.</p>

<p>iljets and quesce...thanks so much!!!</p>

<p>iljets.. are you sure you are correct for part A??? from the way quesce described it, it seems more detailed than what you did.</p>

<p>iljets didn't resolve vectors for part a..there's a horizontal and vertical component that you have to take into account since it's not moving in 1D. It's a 2-dimensional movement.</p>

<p>I don't know what he did for part b and part c - it's kinda messy and I don't want to plough through it, but try to do the question for yourself and see if your answer matches his. If not, I'll give it a go. :)</p>