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 (|$$\delta$$Vy |) 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.

 

[tomcue]

a) The distance from the table top at which we should place the center of the can can be found by setting the y-position equal to the height of the can (h) and solving for x:
x = (S^2/g)*sin(2θ) - (2h/g)*cos^2(θ)

b) The time it will take the ball to reach the top of the can can be found by setting the y-position equal to the height of the can (h) and solving for t:
t = (S/g)*sin(θ) + sqrt((S^2/g^2)*sin^2(θ) + (2h/g))

c) The speed of the ball when it reaches the top of the can can be found by taking the magnitude of the velocity vector at the time t found in part b:
v = sqrt(S^2 + 2gh)

d) The maximum fractional error in the measurement of S can be found by setting the distance from the table top (x) equal to the distance from the table top (x) plus the diameter of the can (D) and solving for S:
S = (g/2h)*(D + sqrt(D^2 + 4gh))

e) The maximum vertical deviation in the velocity vector can be found by considering the maximum possible error in the angle θ, which would result in the maximum possible deviation in the vertical component of the velocity. This can be found by taking the derivative of the vertical position equation with respect to θ and setting it equal to 0:
d/dθ (y(t)) = (S/g)*cos(θ) - (2h/g)*sin(θ) = 0
Solving for θ:
θ = tan^-1(S/(2h))

The maximum vertical deviation in the velocity vector is then given by:
|δVy| = S*sin(θ) = S*sin(tan^-1(S/(2h)))

f) The number of collisions the ball will make with the wall of the can before hitting the bottom can be found by considering the distance the ball travels horizontally before each collision and dividing that by the diameter of the can (D). This will give the number of collisions, n:
n = x/D = (S^2/g)*sin(2θ) / D

Note: These equations assume that the ball is launched horizontally from the table top and that the can is placed directly under the