All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Bouncy Bouncy part 2 (Posted on 2006-02-10)
Part 2: Super-ball

A small ball is thrown toward an incline. This (amazing) ball bounces perfectly.
It is thrown to the right horizontally from the point (0,5) and follows the path of the parabolic equation y=5 - (x^2)/18.
The incline follows the equation y=x/2.

Find the equation of the path the ball takes after its first bounce off of the incline.

(If you wish, find some of the subsequent bounces.)
(Assume that gravity pulls straight down on the ball. Also assume the ball is not spinning so that it is perfectly reflected.)

 See The Solution Submitted by Jer Rating: 4.5000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 solution -- barring any mistakes along the way | Comment 4 of 14 |
(In reply to re(2): potential path (spoiler) by Charlie)

Infinitesimally before the bounce, the ball's trajectory had a slope of -2/3 and infinitesimally after, it had a slope of 18. Taking the ball's speed at the bounce as 1 unit, just before the bounce the x component was cos(arctan(-2/3)) = .8320502943378437, while the y component was sin(arctan(-2/3)) = -.554700196225229. After the bounce the x component was cos(arctan(18)) = 0.05547001962252292 and the y component was sin(arctan(18)) = .9984603532054124.

The height above a given point that the ball will reach is proportional to the square of the vertical component of the velocity at that point. The ratio of the absolute values of the vertical components is .9984603532054124/.554700196225229 = 1.8. Since the vertical velocity just after the bounce is 1.8 times that before the bounce, it will go 1.8^2 = 3.24 times as high.  Before the bounce, the height reached was 5 - 3 = 2 higher than the point (6,3); the maximum after the bounce will be 3.24 * 2 = 6.48 higher than (6,3), or y=9.48.

The time taken to reach this maximum is merely the ratio of the vertical speeds, or 1.8 times as long. But the x-velocity is 0.05547001962252292/.8320502943378437 what it was before, or 1/15, so the distance traveled horizontally is 1.8 / 15 = 3/25 what it was before. The previous x distance was 6 from the crest to the bounce, so now it is 6 * 3 / 25 = 18/25 = .72.  So the apex of the parabola is at (6.72, 9.48) and goes through (6,3).

y = 9.48 - a (x-6.72)^2

and substituting point (6,3),

3 = 9.48 - a .72^2
a = 6.48 / .72^2 = 12.5

making the equation

y = 9.48 - 12.5 (x-6.72)^2

 Posted by Charlie on 2006-02-11 15:48:10

 Search: Search body:
Forums (0)