A stone is thrown vertically upwards and it reaches a maximum height of 30m. What was its initial velocity?

A stone is thrown vertically upwards. So that the only force acting on the stone is the Gravitational force . we consider the the upward directions as positive y-axis . Then there is no motion is along x-axis. And the acceleration of the stone is -g

        where
                    g is  acceleration due  to  gravity

the negative indicates that the acceleration is along negative y-axis (towards downward direction).

The equation of motion connecting the velocities and the displacement of the particle is given by

#v^2=u^2+2*a*S#

Here
v is the final velocity,
u is the initial velocity( velocity with which the stone is thrown)
a is the acceleration . here (-g, acceleration due to gravity),
S is the displacement . here(Maximum Height reached by the stone)

  At the maximum height  velocity becomes zero  ( v=0)

SUBSTITUTE VALUES IN THE EQUATION

#0=u^2-2*g*H#
#u^2=2*g*H#
#u=sqrt(2*g*H)#
#u=sqrt(2xx9.8xx30#
#u=24.248# #ms^-1#

Therefore the velocity with which the stone thrown is #24.248ms^-1#