Derivation of a cinematic formula

[sitzpillow]

I'm really happy about every hint!

Homework Statement

There is a ball on height h which is dropped on a table.
With every Impact the ball loses velocity v by a factor a<1.
I Need to Show the following:
The time T after the ball stopps bouncing is:
T = \frac{1+a}{1-a} (\frac{2h}{g})^\frac{1}{2}

There is a hint to start to calculate the time t_k between the takeoff and the next Impact
and that I should use the geometric series.
We don't need to think about air resistance.

2. The attempt at a solution
My Problem is that if I calculate t_k, I don't include the first time t_0 when the ball gets dropped.
t_k = 2\frac{h_k}{v_k}=2\frac{h_k}{a^k v_0}=a^{-k} (\frac{2h_k}{\sqrt{(2 h_0 g)}}) = a^{-k} (\sqrt{(\frac{4h_k^2}{2 h_0 g}})
with v=\sqrt{2gh} which I got from E_pot=E_kin.
But now I can't shorten h_0 with h_k and a^-k <1 so I can't use the geometric series?!
Is it wrong to use v=s/t ⇔t=s/v since we have a accelarated movement.

 

[Delta2]

Yes its wrong to use that. You should take instead t_k=2\frac{v_k}{g}, k\geq 1, t_0=\sqrt\frac{2h}{g}.

 

[sitzpillow]

How does the v_k looks like?
t_k=2 \frac{a^k v_0}{g} = 2 a^k \frac{\sqrt{2gh}}{g}= 2 a^k \sqrt{\frac{2h}{g}} ?
What's next?

 

[Delta2]

Next is geometric series of $t_k$. $T=t_0+\sum\limits_{k=1}^{\infty}t_k=\sqrt\frac{2h}{g}+\sum\limits_{k=1}^{\infty}t_k$. Notice that the actual $t_0$ is half of what you get from the formula of $t_k$ for k=0(because we drop the ball from above the table), that's why i put it outside the summation which summation is done from k=1 for this reason.

or you can do the series from k=0 but then subtract $\sqrt\frac{2h}{g}$.

 

[sitzpillow]

Thank you. To finish that threat I conclude:

t_k means the time between the k and k+1 impact.

for t_0
we get
t_0=\sqrt{\frac{2h}{g}} with s=ut+\frac{1}{2}gt^2

for t_k with k>0
\Sigma{t_k}=\Sigma_1^n{2 \frac{a^k v_0}{g}}=\Sigma_0^n{2 \frac{a^k v_0}{g}}-\frac{2v_0}{g}
now we let n go to infinity and use the geometric series:
\Sigma_{n=0}^{\infty}{t_k}=\Sigma_{n=0}^{\infty}{2 \frac{a^k v_0}{g}}-\frac{2v_0}{g}=2\frac{v_0}{g}(\frac{1}{1-a}-1)=\frac{v_0}{g}\frac{2a}{1-a}=\sqrt{\frac{2h}{g}}\frac{2a}{1-a}
by using E_{kin}=E_{pot}

Adding 1. plus 2. we get: \sqrt{\frac{2h}{g}}\frac{1+a}{1-a}

 

[haruspex]

There is a slightly easier way.
Consider the ball starting by rising from the ground at speed v. The time to first bounce is 2v/g. If the time to coming to rest is t(v) then t(v) = 2v/g + t(av). Guess a linear solution, t(v) = kv, and find that $k=\frac{2}{g(1-a)}$. Then it is a matter of converting to the given initial conditions.