Astone dropped from the root of a high buliding . a second stone is dropped 1.30 s later how far apart are the stones when the second one has reached a speed of 14.0 m/s

The stones are apart one each the other 26.37 meters when the second one has reached a speed of 14 m/s.

Explanation:

Second Stone:

Vb= g*t

tb (14m/s)= Vb/g

tb (14m/s)= (14 m/s) / (9.8 m/s²)

tb (14m/s)= 1.42 sec

Hb= (g * tb²)/2

Hb= 9.88m

First Stone:

ta= tb + 1.3 sec

ta= 2.72 sec

Ha= (g*ta²)/2

Ha= 36.25m

Distance between Stones:

Ha-Hb= 36.25m - 9.88m

Ha-Hb= 26.37m

The photon striking the atom is moving too fast.

Explanation: