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If we throw a ball from the hight point $h$ from the earth, with initial velocity $v’$, how to prove that the time it takes the ball to reach the earth is given by:

$$t=\frac{v}{g}(1+\sqrt{1+\frac{2hg}{v^2} } )$$

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closed as off-topic by DavePhD, Qmechanic May 10 '14 at 14:08

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – DavePhD, Qmechanic
If this question can be reworded to fit the rules in the help center, please edit the question.

You work with the equations of motion. BTW, is this a homework type problem? – ja72 Sep 26 '12 at 13:32
Why the question was tagged as a homework ? It's not – rib Sep 26 '12 at 14:56
Did you check the homework tag description? – Qmechanic Sep 26 '12 at 15:38
@Qmechanic Sorry no,but i know that you shouldn't tag the question as a homework unless if the OP say that blatantly. – rib Sep 26 '12 at 15:47
@rib This won't work -- hard core do-my-job OPs will never admit their question is a homework problem. – mbq Oct 7 '12 at 11:35

For a free falling object without air resistance you have two equations

$$ y = h + v'\,t - \frac{1}{2} g t^2 $$ $$ v = v' - g\,t $$

with $h$ the initial height, $v'$ the initial velocity (upwards is positive), $y$ the height at time $t$, and $v$ the velocity.

Solve them when $y=0$ for $v$ and $t$.

Reference: projectile motion.

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equations should be $ v_{f}=V_{0}+gt $ and $ h=v_{0} t+1/gt^{2} $ the equation for the high is a second order equation yu can solve it immediatly for 't' – Jose Javier Garcia Sep 26 '12 at 15:02

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