The equation for the first ball dropped under the force of gravity is s=gt^2/2 where g is the acceleration of gravity, s is the distance=h/2 and t the time taken to fall the distance s. For the second ball, s=vt-gt^2/2, where v is the vertical velocity which needs to overcome gravity. h/2=gt^2/2, so t=sqrt(h/g). Putting this value of t into the other equation we get h/2=vsqrt(h/g)-g*h/2g. vsqrt(h/g)=h and v=h/sqrt(h/g)=sqrt(h^2g/h)=sqrt(hg). The value of g is 32'/s^2 so v=sqrt(hg)=4sqrt(2h), where h is in feet.