# Bullet fired at a series of partitions

Imagine a bullet fired at a series of partitions stacked one after the other. Given that the bullet looses half its velocity in crossing each partition, velocity of the same is a geometric progression(GP) with $a= v_0$ and $r=0.5$.

The question is, how many partitions is it supposed to successfully cross, or at what partition would the bullet be stopped?

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There's something missing from the question. What's the minimum velocity needed to break through a single partition? –  EnergyNumbers Apr 2 '12 at 12:03
Are you trying to recreate Zeno's paradox? –  Alexander Jun 1 '12 at 11:07

Looks like the answer is $\infty$/"never", since the velocity after the $i^{\rm{th}}$ partition is $v_i=ar^{i-1}$ by normal GP formulae; and an exponential can only approach zero. But you probably knew that.
@Vaibhav No, you'll get an infinite series there as well. The common ratio will be $\frac34$ instead. Are you sure that the fraction $\frac12$ was of the velocity before hitting the plate or of the initial velocity? In the latter case the problem is trivial, answer is 2 (you get an AP, not a GP) –  Manishearth Apr 2 '12 at 7:53
@Bernhard: Uhh, how? Its getting slower, not faster. If it gets slower, it will take more and more time to reach the next barrier. So the time taken will be $\sum\limits_{i=0}^\infty \frac{d}{v}=\sum \frac{d\times \Large{2^i}}{v_0}=\infty$ –  Manishearth Apr 2 '12 at 8:14