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.

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:


v_t = \sqrt {\frac {2 W}{\rho A C_d}}


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:


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


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.