How to Calculate the Ball's Trajectory to Land in the Can?

[2011vlu]

Homework Statement

A ball rolls off the end of a table of height H. The ball has speed S when it leaves the horizontal table top. We wish to place a can of height h and diameter D at the correct distance from the table’s edge so that the ball lands in the can (aim for the top center of the can!) Find expressions for the following in terms of any/all of H, h, D, S, and g.

a) The distance from the table top (x) at which we should place the center of the can.
b) The time it will take the ball to reach the top of the can.
c) The speed the ball will be moving when it reaches the top of the can.
d) The maximum fractional error in our measurement of S which is allowed so that we will still get the ball in the can (assuming perfectly horizontal launch from the table top).
e) The maximum vertical deviation (|\deltaVy |) for the velocity vector as the ball leaves the table top that is allowed if we are to still get the ball in the can. Assume that the speed of the ball as it leaves the table top is still S.
f) Assuming a perfect insertion into the middle of the top of the can and that the effect of a collision with the walls of the can is "perfectly elastic"; meaning that the effect of a collision with the wall of the can is to instantaneously reverse the horizontal component of the ball's velocity, develop an expression for the number of collisions the ball will make with the wall of the can before it hits the bottom of the can (where there is some sticky stuff which stops the ball instantly).

Homework Equations

x(t)=x0+(v0cos\theta)t
y(t)=y0+(v0sin\theta)t-(1/2)gt^2

The Attempt at a Solution

a) I solved for t when y(t) is h because that's how tall the can is. I got x(t)=S(rad((2H-2h)/g))
b) I already solved for t in the above section, and I got t=rad((2H-2h)/g)
c) I found the derivative of the position to get the velocity, found the magnitude of that vector, and got rad(S^2-2gH-2gh)
d) I'm stuck, I THINK what I'm supposed to do is S(rad((2H-2h)/g))-D=x(rad((2H-2h)/g)), but it doesn't quite work.
e) Please help with this part.
f) Please help with this part.

 

[Delphi51]

c) I found the derivative of the position to get the velocity, found the magnitude of that vector, and got rad(S^2-2gH-2gh)

v = Vi - g*t = -g*t (no S involved in the vertical motion!)

For (d) I would start with the extra speed needed to hit the far edge of the can divided by the center speed.

(delta S)/S = (x + D/2)/t - x/t all divided by x/t
it works out very simply to .5*D/x as it should since everything is linear.

I played with (e) a bit. I used Sx and Sy for the speed components and worked the problem again with horizontal distance x + D/2. Got an expression for Sy (which is the same as delta Sy) that has Sx in it a couple of times. Could get rid of that using Sx^2 + Sy^2 = s^2.
Pretty messy. I'm sure you have to work the problem with an initial vertical velocity but perhaps there is a way to avoid the mess assuming small deviations and using calculus.