
#1
Dec2011, 10:57 AM

P: 13

1. The problem statement, all variables and given/known data
You kick a ball at 20m/s at 40* from the horizontal, at the very top it hits a bird, stops immediately and fall straight down. vi=20m/s θ=40° I need to calculate the velocity right before impact the max height (height of collision) speed of ball when right before it hits the ground 2. Relevant equations i need to relate the velocity at 2 different points and time is unknown so i should use Conservation of energy EfEi=EinEout 3. The attempt at a solution Im not sure what i want to set as my initial and final points, im thinking Ei=1/2mvi^2 Ef=1/2mvf^2 + mgh in and out=0 just want to know if i have this setup properly, I dont have the answers anywhere and they are not written down anywhere. from the energy equation m will cancel out and i will be able to solve for the final velocity in terms of initial velocity and height, then im pretty sure i need to add a second approach to get the height. 



#2
Dec2011, 11:07 AM

P: 1,084

to get the max height all you'll need to do is translate all of the initial kinetic energy to potential energy. You have that equation.




#3
Dec2011, 11:09 AM

P: 13





#4
Dec2011, 11:40 AM

P: 1,084

Kicked ball hits bird and fallsWith this, you should be able to calculate the maximum height (using the y component of the trajectory speed). Once it hits the bird, it falls straight down (in other words, the x component of velocity goes to 0 upon impact). Beyond that, I don't really find it necessary to tell you how to solve the question, as I'm sure there are many ways. HINT: If the impact occurs at the top of the trajectory, the ycomponent of the velocity is 0 at that time. 



#5
Dec2011, 11:45 AM

P: 38

No need to make it so complicated; most projectile questions can simply be solved using the 3 equations.
v = u + at s = ut + 1/2 at^2 v^2 = u^2 + 2as All you need is to think about the initial condition, and final condition.  To find the velocity at the very peak (i.e. right before impact), just think about the motion of the ball. What components does the velocity have at this point in time, and how does this relate to the initial condition? (You don't even need the equations of motion for this one!)  For the max height, you can as dacruick said use the conservation of energy, but you could just again use the equations for motion. Think about the motion of the ball at its max height. What is its velocity, acceleration, etc. The question asks for max height, so this implies that you would apply the chosen equation in the y direction.  Speed of the ball before it hits the ground is kind of dodgy, since the bird influenced the ball and there would have been a collision. I'm not really sure what your teacher wants. I guess you still just assume conservation of energy applies to the ball and ignore the bird =\ 



#6
Dec2011, 12:35 PM

P: 13

ok so here is what i did:
y: h=g/2 (t)[itex]^{}2[/itex] + V[itex]_{}i[/itex]Sinθ(t) ΔV[itex]_{}y[/itex]=a[itex]_{}y[/itex](t) t=(V[itex]_{}i[/itex]Sinθ)[itex]\frac{}{}g[/itex] t=1.31 seconds h=25.25 meters (makes sense since it was going 20 m/s and traveled for a bit over 1 second) now for the final velocity, since the y component of velocity will be 0 at the peak, i did V[itex]_{}f[/itex]=V[itex]_{}i[/itex]Cosθ Vf=15.32 m/s the speed after impact (assume energy is conserved and the bird didnt flail :P) v=√(2gh) v=22.25 m/s all these answers do not seem unreasonable, did i do it right? 



#7
Dec2011, 12:49 PM

P: 1,084

I'm not sure which questions you are answering.
Hitting a bird and stopping doesn't really scream "conservation of energy" to me. How can the final velocity be higher than the initial velocity, where would that extra energy come from, the bird?? It seems like you're doing a lot of equation searching, and then a plug and chug strategy. is this the case? I strongly advise against that, especially since almost every energy question I've done can be simplified by understanding the concepts. as clementc suggested, you can use the initial conditions and final conditions to isolate the things you're solving for. I think it would be a more useful approach for you to explain the reasoning behind each prospective calculation. 



#8
Dec2011, 01:59 PM

P: 13





#9
Dec2011, 02:04 PM

P: 1,084

I am able to do all of your problems without using a single equation of motion, yet you have a bunch of them. What I mean to do is to make you aware of the essence of these energy calculations, as well as help you practice logically working through an answer. 



#10
Dec2011, 02:06 PM

P: 1,084

Let me give you an example.
The first question asks you for the velocity right before impact, which is said to be at the top of the trajectory. I know that at the top of the trajectory, the ball is stationary in the ydirection. Therefore the velocity will be only in the xdirection, and it will be equal to the initial xcomponent of velocity (which is 20 m/s * cos(40)). EDIT: I know that you already have solved for that part of the question, I am just saying that there are similar simplifications that can be made to solve the other two. Once you know how to do this, it might be worth your time to double check them using the formulas. I'll also point this out again. I do not believe energy is conserved upon impact with that bird. 



#11
Dec2011, 02:21 PM

P: 13

you get 80% credit for the problem as long as you have the correct equations setup and solved for the correct variable. and yes, energy is definitely not conserved when the bird was hit but we are asked to assume it is conserved. 



#12
Dec2011, 02:36 PM

P: 1,084

I don't understand how energy can even be assumed to be conserved upon impact with the bird. Especially since the question states that the ball drops immediately after the impact, strongly suggesting an inelastic collision. However, I do sympathize with your exam being tomorrow, and I understand that it cannot always be about the deep understanding, sometimes you just have to get sh*t done haha. So in light of that, your answer to the first question is correct at 15.32 m/s (but the question asks you for velocity, you have just stated a speed. Don't forget to add that it's in the xdirection). The second question asks you for the maximum height. How I would go about this is as follows: We know that the xcomponent of velocity does not play a role in the height. You've chosen to use a kinematic equation, and have ended up with a peculiar answer of 25.25 metres. At the maximum height, all of the initial kinetic energy from the ycomponent of velocity has been transferred to potential energy. Therefore you should have a simple mgh = 1/2 m (Vinitial)^2, in which you could easily solve for h. Do you understand this? Make sure that you know that the initial velocity is only the ycomponent As for the third question, which is apparently up for debate, I am going to assume it is an inelastic equation because I do not see a way to proceed in the case that energy is conserved. It asks you to find the speed right before it hits the ground, and these are the conclusions I will make before I try to calculate The ball collides with the bird when the ycomponent of velocity is 0 The ball loses all xcomponent velocity upon impact (it falls straight down) From these conclusions, I can only think one thing. This question could be simplified to, "What is the speed of a ball being dropped from the maximum height (from question 2) at the instant before it hits the ground." Do you have any qualms with that interpretation of the question? 



#13
Dec2011, 03:29 PM

P: 13

yes, that makes sense to interpret it like that.



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