I think it will end up at (D,-h).Originally Posted by dopi
Let be the total time of flight.
You can use your equation:Y = Vo*sin(alpha)*t - 1/2*g*t^2
If t= , then Y=-h.
Rest you can do yourself.
a tennis ball is hit from a height h above the ground with a speed V and at an angle alpha to the horizontal. it hits a wall at a horizontal distance d away . Air resistance is negligible
i was told to find expressions for the height and velocity of the ball for which i did shown below
Vx = Vo*cos(alpha)
Vo is the initial velocity
alpha is the launch angle
Vx is the x component of velocity
Vy = Voy*sin(alpha) - g*t
From the pythagorean theorem V = sqrt(Vx^2+Vy^2)
X = Vo*cos(alpha)*t
Y = Vo*sin(alpha)*t - 1/2*g*t^2
Assuming that i started at (x,y) = (0,0) i will then end upat the point (D,h)
that was the first part of the quesiton...but the part im stuck on is ...auusuming that the coefficient of restitution between the tennis ball and the wall is e, i need to write down an expression for te the velocity of the ball immediately after impact......is there anyone that can help ???thankz
i dont understand how that will help me to find an expression for the velocity of the ball immediately after the impact, when the coefficient of restitution between the ball and the wass is e.????Originally Posted by malaygoel
You have done good work, I will use two of your equations(see quote).Originally Posted by dopi
To find the velocity after impact, we need to know two things:
coefficient of restitution(which we know)
velocity before impact(which is our focus now)
Velocity before impact can be divided into two components:
horizontal component(which you know )
vertical component--- for it you need to know the total time of flight[t1](since it depends on t)
You could find t1 using the equation:
X = Vo*cos(alpha)*t
I hope you have got your answer. Feel free to say.
The definition for the coefficient of restitution is:Originally Posted by dopi
So if you know the KE before collision (KE0) you can find the KE after collision (KEf). This also works for multibody collisions where the KE's in the formula are the total KE for the system.
By the way, you say you are looking for the velocity after collision. That isn't correct. You can only get the speed after collision using this equation. What's going to happen to the velocity is anyone's guess because we don't know what effect the collision will have on the individual components of velocity. (The only example I can think of where you CAN get the final velocity is if the collision is "head-on." Then the velocity after impact will be directly opposite the velocity before impact.)
Well, that's how I learned it anyway, I think. (Sigh) I went back to the book I thought I had gotten it from (Serway's intro Physics book) and couldn't even find it! Seeing as I can't find it in my Graduate Mechanics book either, I obviously learned it from some out of book notes. I could simply be remembering it wrong.Originally Posted by malaygoel
I propose we do an experiment. Let us throw a ball gently on the patio soOriginally Posted by CaptainBlack
it bounces and has forward motion. If Topsquark is right the ball will stop
bouncing with no residual velocity along the patio. If I am right it will continue
rolling along the patio with approximately the same speed as the original
horizontal component of the velocity (some loss to friction and conversion
of translational to rotational energy), yes?
I've done this experiment, and so know the answer
PS my spell checker wants to change TopSquark to pipsqeak!
The point is, using my definition, the ball would eventually stop even over a frictionless surface. (e is always less than 1 in reality) If the ball had a horizontal component of velocity Netwon's 1st says it should always be in motion. A contradiction, so my definition can't be correct.Originally Posted by malaygoel
Yes. According to CaptainBlack (and I no longer have any reason to doubt his definiton) the coefficient of restitution is a ratio between the final and initial "vertical" components of the velocity. (Vertical referring to a normal to the surface of impact.)Originally Posted by malaygoel