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oonej
Joined: 20 Apr 2011 Posts: 2 Location: Tampa, FL

Posted: Wed Apr 20, 2011 4:41 pm Post subject: Bouncing Ball with a COR 


Hey guys,
I just wanted to mention that I have posted this same exact question on numerous websites with little to know success. I hope we can figure this out together!
Consider there is a window 5 meters from a person. The window is sized 1.21 meters (4 ft) in length and .914 meter (3 ft) wide at a height of 6 meters from the ground.
A person 1.8288 meters tall (6 ft) wants to bounce a ball (COR: 0.789) on the ground such that the ball goes through the window on the first bounce.
If the person bounces the ball with a speed of 25m/s at an angle of 60° from the horizontal at a point in the middle (between him and the wall), calculate and plot the trajectory of the ball.
Will the ball go through the window? If not, change one of the values such that it does. Then find a RANGE of velocities that will meet this condition, while keeping the COR, Distance, and angle the same.
I have discovered that this does not actually work. So according to the problem, before we proceed, we must make the necessary corrections. Which I have discovered that changing the angle to 73 degree's, in fact fixes our problem. The ball clears the window at 6.5 meters in height. Which is almost directly in between the lowest and highest possible height.
My problem lies on the second part. Finding the range of velocities that will make this work. This seems to be the part that everyone gets lost on. I was given a hint from my professor to combine the x(t) and y(t) functions into a function of y(x) which I have done. But I get ridiculously low velocities which do not work when I plug them back into the equations. 

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hepcj Site Admin
Joined: 23 Jun 2007 Posts: 125

Posted: Thu Apr 21, 2011 12:54 pm Post subject: 


Are you considering the effect of gravity? The initial velocity of the ball 25m/s is quite high compared to the distance travelled. The problem would be greatly simplified by neglecting the parabolic path of the ball and will give a ballpark figure (no pun intended). 

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oonej
Joined: 20 Apr 2011 Posts: 2 Location: Tampa, FL

Posted: Fri Apr 22, 2011 4:45 am Post subject: 


No gravity has to be considered.
And 25 m/s does not clear the window when considering the COR.
So the problem states to change the problem by changing one variable. I chose to change the variable of the angle to 73 because I tested the speed at all possible velocities. And calculated the trajectory to always impact under the window. The angles (broken down into the triangles for "max velocity" shows that the ball will never clear the window's height.
If you are unsure about the COR it changes the trajectory angle after every bounce. Thus making it project at 53 degrees after the first bounce... making it impossible to make it to the height of 6 meters based on the ~4 meters distance it has to travel after the first bounce.
I haven't had much success with this problem...
Asked every where and everyone i know that has a lick of knowledge with physics.... 

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hepcj Site Admin
Joined: 23 Jun 2007 Posts: 125

Posted: Mon Apr 25, 2011 2:31 am Post subject: 


Okay, lets make sure we are singing from the same hymn sheet.
As I read the question it is saying change one of the variables such that it will go through the window and in this state then change the velocity so that it clears the bottom of the window and then increase the velocity so that it will reach the top of the window. This will give you a range of velocities in which it will go through the window.
First I calculated the components of velocity vx0 and vy0 from the angle theta and the velocity of 25 m/s. Next I calculated the velocity at some position x where the ball hits the ground. The xcomponent of velocity is unchanged but the ycomponent increases slightly due to gravity. I then use
height of person = vy t  0.5 g t^2 and insert values for t as a function of x. i.e t = x/vx0
Now I can find the position at which the ball hits the ground and I know the incident velocity. vy1 and the angle theta_i from the normal using Arctan(vy1/vx0)
The reflected angle is calculated from the yvelocity of the rebound. Which using the c as the coefficient of friction. vy2 = vy1 c
The xcomponent is still unchanged. The reflected angle is then Arctan(vy2/vx0) although we don't actually need this.
We can calculate the height that the ball reaches using
h = vy2 t  0.5 g t^2 only this time t = (5x)/vx0 because the distance to the wall is 5 metres.
Doing this I changed the angle to 63.5 deg and this gives a height of 6.05 m.
I can increase the velocity and the height increases slightly but it will always go through the window. The method seems right but the idea that ball will always go through the window at any velocity above 25m/s is a bit dubious. 

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