Question: Man jumps from an airplane at an altitude of h metres. Determine the impact velocity upon landing. Express it in terms of $h$.
$$v = \sqrt{5g} \cdot \tanh\left(\sqrt{\frac{g}{5}}t\right)$$
Attempt to solve:
I first integrate it to get:
$$h(t) = 5\sqrt{g} \cdot \ln\left(\cosh \left(\sqrt{\frac{g}{5}}t\right)\right)+C$$
Then from the question, I got the condition h=h when $t=0$, thus:
$$h(t) = 5\sqrt{g} \cdot \ln\left(\cosh \left(\sqrt{\frac{g}{5}}t\right)\right)+h$$
Then I am stuck. I think impact velocity is velocity when $h=0$ but I am not sure.
Do tell me if information is not sufficient to solve this problem.
Edit: Velocity at time t with the downwards direction is taken as positive.
The velocity is toward the ground so if you're defining $h(0) = h$ and that impact is when $h(t) =0$ you want to use the equation $$ h(t) = h-5\sqrt{g}\ln\left(\cosh\left(\sqrt{\frac{g}{5}}t\right)\right)$$ for the height.