A stone is dropped from the roof of a high building. A second stone is dropped 1.00 s later. How far apart are the stones when the second one has reached a speed of 14.0 m/s?

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