Answer:
Velocity v= 12.25 [tex]\frac{m}{s}[/tex]
Explanation:
The first rock dropped give the distance Y in meters
[tex]Y_{f}=Y_{o}+v_{o}*t +\frac{1}{2}*a*t^{2}\\ Y_{f}=\frac{1}{2}* 9.8 \frac{m}{s^{2} }* 3^{2} \\Y_{f}=44.1 m[/tex]
Now the motion of the second rock the time change so to know the velocity
[tex]Y_{f}= Y_{o} +v_{o}*t +\frac{1}{2}*a *t^{2} \\v_{o}*t= -Y_{f} +\frac{1}{2} *a*t^{2} \\v_{o} =\frac{-44.1 +0.5 * 9.8* s^{2} }{2} \\v_{o}= 12.25 \frac{m}{s}[/tex]
