Two baseballs are fired into a pile of hay. If one has twice the speed of the other, how much farther does the faster baseball penetrate? (Assume that the force of the haystack on the baseballs is constant).

When there is a two-fold increase in speed, there is a four-fold increase in stopping distance. For constant resistance forces, stopping distance is proportional to the square of the speed. 

$\text{Stopping distance } S \propto u^2$

$\text{If speed is doubled then } S'\propto 4u^2 $

$\Rightarrow \frac{S'}{S} = \frac{4u^2}{u^2} = 4$

$\Rightarrow S'= 4S$