An ball is dropped from 20.5m. It bounces once, falls back down, then bounces a second time to 7.6m. What's the max height of the first bounce?
An ball is dropped from 20.5m. It bounces once, falls back down, then bounces a second time to 7.6m...
Carson Cruz
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Parker Ross
Depends on the assumptions you make.
Ryan Sullivan
>an ball
Dominic Rogers
Not on earth it doesn't.
Ayden Allen
gh=10*20.5=205 m^2/s^2
gh'=10*7.6=76 m^/s^2
205-76=129 m^2/s^2
/10m/s^2 = 12.9m
Wyatt Anderson
>205-76=129 m^2/s^2
>/10m/s^2 = 12.9m
What the fuck are you doing?
Isaiah Harris
assume the energy lost is proportional to total energy by a factor of x
then 20*x^2=7
20*x =h , h being what we want to compute
7*20 =h^2
√(140) =h
h≈12m
Carson Robinson
also dude golden ratio lmao
Dylan Gutierrez
Physics.
Jaxson Moore
For some strange reason you have assumed that the ball has energy after the first impact equal to the difference between its initial energy and energy after the second impact. No basis for this and it makes no sense.
Thats a big assumption, but maybe needs to be assumed.
Adam Kelly
>Physics.
No you're not, you're banging numbers together and hoping you get an answer.
>Thats a big assumption, but maybe needs to be assumed.
It seems like a pretty reasonable assumption to me.
Thomas Hernandez
Seems like a pretty convenient assumption to me. With no knowledge of the ball you have no idea if its true, and without it you wouldnt be able to solve the problem with this method.
Mason Ross
If you have some better way of showing it, please do. Your answer will be between 12 and 13. Otherwise you have no idea what you're doing, which is what i suspect.
Hunter Taylor
States:
1: upon releasing the ball
2: instantaneously before impact
3: next max ascent
4: instantaneously before impact
5: next max ascent
[eqn] E_1 = mgh_1 = E_2 = 1/2 mv_2^2 [/eqn]
[eqn] E_3 = E_1 - X_{2-3} = mgh_3 = E_4 = 1/2 mv_4^2 [/eqn]
[eqn] E_5 = E_4 - X_{4-5} = mgh_4 [/eqn]
idk
Kayden Russell
Assuming exponential decay after each collision, I got 12.48m.
Ayden Ward
You have no model of the collision, so there's no way to really know.
I put the upper bound at 20.5 meters.
Logan Morris
What's 'exponential decay'?
James Rodriguez
I suppose that the general energy drop follows an exponensial form.
Jeremiah Barnes
Samuel Ortiz
12.5m
Anthony Myers
best solution so far
apply principle, use 'ruthless rounding', get sanity check for detailed calculation.
popular exam question: will it bounce forever? if not, how long?
answer: the number of bounces is not limited but the time is.
(in OP's case it would bounce for about 16 seconds)