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malik20 · Oct 16, 2011, 10:12 AM

How high do we throw a cricket ball so that when it hits the ground, it has no acceleration? Its drag coefficient is 0.5.

ebaines · Oct 20, 2011, 08:12 AM

You have to throw it infinitely high. This is because the equation of a falling object's velocity versus time assymtotically approaches the "terminal velocity," which is the theoretical maximum speed the object reaches as it falls. Terminal velocity is relatively easy to calculate:

$<br /> v_t = \sqrt {\frac {2 W}{\rho A C_d}}<br />$

where W is the weight of the ball, $\rho$ is the density of air, A is the ball's cross-sectional area, and $C_d$ is the drag coefficient. But the velocity never quite reaches this limit. Mathematically the velocity equation is:

$<br /> v(t) = \sqrt { \frac {2 W}{\rho A C_d}} tanh \left( t \sqrt{\frac {g \rho C_d A}{2m}} \right)<br />$

The tanh() function approaches 1 for large values of t but never is actually equal to 1. See the figure below for an example of how this function behaves. So while v(t) comes very very close to $v_t$ , it never quite makes it. Hence there is always a small bit of acceleration still happening no matter how far the ball drops.