Shayaan_Mustafa
  • Shayaan_Mustafa
A ball is thrown from height h vertically upward with initial velocity Vo. Find the final velocity of ball with it hits the ground?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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TuringTest
  • TuringTest
if it has velocity \(v_0\) on the way up, then when it gets to height h it will have the same velocity again only downward, right?|dw:1332798412849:dw|
Shayaan_Mustafa
  • Shayaan_Mustafa
yes right.
TuringTest
  • TuringTest
so we have the problem\[v_f^2=v_0^2+2ad\]in this case this is\[v_f^2=v_0^2+2gh\implies v_f=\sqrt{v_0^2+2gh}\]or if you want to not the fact that we are going in the negative direction\[v_f=-\sqrt{v_0^2+2gh}\]

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TuringTest
  • TuringTest
want to note the fact that*
Shayaan_Mustafa
  • Shayaan_Mustafa
\[h=V _{o}^{2}/2g\]am I right?
TuringTest
  • TuringTest
yes
TuringTest
  • TuringTest
ok I see where you got your formula now
Shayaan_Mustafa
  • Shayaan_Mustafa
so\[v _{f}=-\sqrt{v _{o}^{2}+v _{o}^{2}}\]is it correct??
TuringTest
  • TuringTest
but we are confusing two things, I was mistaken to say you were right that\[h=v_0^2/2g\]you are talking about the \(maximum\) height in that equation h for us is unknown
Shayaan_Mustafa
  • Shayaan_Mustafa
then answer is\[v _{f}=-v _{0}\]
Shayaan_Mustafa
  • Shayaan_Mustafa
yes. ball goes to maximum height.
TuringTest
  • TuringTest
|dw:1332799001302:dw|let us call the \(maximum\) height it reaches H, we then have\[H={v_o^2\over2g}\]so if you want to sub that into \[v_f=-\sqrt{v_0^2+2gh}\]that will not work because the inital heigh h is not the same as the maximum displacement from h, which is H the maximum height it reaches would be h+H
TuringTest
  • TuringTest
typo, the maximum height it reaches is not H H is the \(additional\) height the ball reaches from its initial height h
Shayaan_Mustafa
  • Shayaan_Mustafa
yes yes. exactly . I also did this. now I want to move a step ahead.
Shayaan_Mustafa
  • Shayaan_Mustafa
yes..
TuringTest
  • TuringTest
I think as far as we can take it with these variables is we have\[H={v_o^2\over2g}\]\[v_f=-\sqrt{v_0^2+2gh}=-\sqrt{v_0^2+{v_0^2h\over H}}=-v_o\sqrt{1+\frac hH}\]but we didn't know H to begin with, so unless that is given in the problem I don't think this is a very useful formula.
Shayaan_Mustafa
  • Shayaan_Mustafa
yes turning test. I also got same formula. and my expressions are same as your. well thnx for help buddy.
TuringTest
  • TuringTest
anytime bro!

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