A rock is thrown upward with an initial velocity of 9.6m/s to what hight does it rise and how long does it take to reach that high

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