We know that the first stone accomplish the rule : Vf=Vi+a*t
if we consider that initial speed is 0 m/s and the acceleration of the gravity is 9.8m/s2
then we can calculate the time elapsed when the stone reach 14 m/s, that is:
14 = 0 + 9.8*t => t = 14/9.8 = 1.43 seconds
Then the distance dropped within that time can be calculated:
d = Vi*t + 1/2*a*t^2 = 0*1.43 + 1/2*9.8*1.43^2 = 10 metres
For the second stone we can apply the same calculations, taking in account that is dropped 1 second later, so the time elapsed will be 1.43 - 1 = 0.43 seconds, and the distance will be:
d = 0*0.43 + 1/2*9.8*(0.43)^2 = 0.9 metres
So the distance between both stones will be: 10 - 0.9 = 9.1 meters
