What is your team up to at your hotel?

[AdamHeard]

Originally Posted by Otaku

Assuming that the bouncyball weighs a tenth of a pound (or so, .05kg to be precise), dropped from the very top of the atrium to the very bottom...

67.13 Newtons.

That is (assuming I remember correctly) four times the force of a golfball dropped from one meter.

So yeah. It'd hurt. Because golf balls dropped from one meter hurt. But this would hurt more.

EDIT: Whoops! For some reason 40 * 4 = 60 in my brain

but basically it's a lot of force. And would hurt. Even if you take off the 2m or so from the height (assuming it hits a 2m tall person on top of the head) it's still going to hurt.

edit:

that's 2.7m/s, or .1mph.


edit (yes again):

It will take 52 seconds to hit the ground, too.

edit (whee):

I did some more math

here's the full thing:
"A bouncyball dropped from the top of the atrium at the Mariott Marquis, assuming it weighs .05kg, Will hit the ground with a force of 67.13N. The ball falls at a speed of 2.78m/s and will take 52 seconds to hit the ground. Assuming optimal conditions, that's going to be a 132m rebound at 90% efficiency. The ball, dropped from the 47th floor, will get back up to the 42nd floor."

Seems a little off to me....

Acceleration of the ball would be 9.8 m/s/s.

I've never been there.... so 47 floors I'm assuming is 705 ft (15*47), which is ~215 m. So, integrating the velocity shows that it hits 215m at 6.6 seconds.

I'm betting my guesstimation of height is off... but it's a closer approximation of time at least.

I'm too tired to finish.... I'll let Otaku take over from here and come up with the energy; goodnight everyone.


EDIT: forgot terminal velocity. It hits 54m/s after 5.5s and has traveled 148m, leaving 67 to go. So add 67m / 54m/s and you get a total of 6.75 s at 54m/s (my old calc, physics and chem teachers would kill me for all the approximations and random rounding).