Given:
u(initial velocity):9.6 m/s
at the maximum height final velocity =0
The acceleration acting on the body is gravity as it is free falling object
=-9.8m/s^2
Now we know that
v^2-u^2 = 2as
Where v is the final velocity measured in m/s
u is the initial velocity which is measured in m/s
a is the acceleration measured in m/s^2
s is the distance traveled by the rock in this case it is considered as the height
Substituting these values we get
0-9.6= 2 x -9.8 x s
s= 0.49 m
Consider t as the time taken for the rock to travel
v=u+at
0=9.6 -9.8t
t=0.98sec