Can we reach escape velocity ?

$n = \frac{log(\sqrt{R}+1)}{log(2)} \approx 11.3\ balls$
$n = 12\ balls$

For Simplicity, Let us assume there is a very small distance, so that the relative bounces happen after a short time interval. Just before collision with the ground, both balls are moving with same velocity i.e. v = $\sqrt {2gh}$ Just after collision, the basket ball is coming up with velocity v and the upper ball is coming down with velocity v. The relative speed them is therefore 2v. After the balls bounce off each other, the relative speed is still 2v. (This is clear if you look at things in the frame of the basketball, which is essentially a brick wall.) Since the upward speed of the basketball essentially stays equal to v (it is like a big ball with little effect of collision) , the upward speed of the tennis ball is 2v + v = 3v Similarly the velocity of third ball will be (3v + v) + v = 7v. Similarly, the velocity of (i+1)th ball would be $v_{i}$ + v + $v_{i}$ = 2 $v_{i}$ + v. Using concepts of Sequences and Series, we find $v_{n}$ = ( $2^{n}$ - 1 )v

To reach escape velocity, $v_{n}$ >= $\sqrt {2gR}$ . From this we find $\boxed{n = 13}$ . Of course, the elasticity assumption is absurd in this case, as is the notion that one can find 13 balls with the property that $m_1 \gg m_2 \gg \ldots m_{12}$ .

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