A ball is thrown from height h vertically upward with initial velocity Vo. Find the final velocity of ball with it hits the ground?
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|
yes right.
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\[v_f=-\sqrt{v_0^2+2gh}\]
want to note the fact that*
\[h=V _{o}^{2}/2g\]am I right?
yes
ok I see where you got your formula now
so\[v _{f}=-\sqrt{v _{o}^{2}+v _{o}^{2}}\]is it correct??
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
then answer is\[v _{f}=-v _{0}\]
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