# Proving $t=(1+\sqrt{1+2hg/v^2 } ) (v/g)$ for a thrown ball

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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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
– 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
I'm just trying to get a decision either way. If you think the homework tag does not apply, please edit it out again. – Qmechanic♦ Sep 26 '12 at 16:22
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## 1 Answer

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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