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

Sep 17, 2017

$24.248 m {s}^{-} 1$

#### Explanation:

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

${v}^{2} = {u}^{2} + 2 \cdot a \cdot S$

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

$0 = {u}^{2} - 2 \cdot g \cdot H$
${u}^{2} = 2 \cdot g \cdot H$
$u = \sqrt{2 \cdot g \cdot H}$
u=sqrt(2xx9.8xx30
$u = 24.248$ $m {s}^{-} 1$

Therefore the velocity with which the stone thrown is $24.248 m {s}^{-} 1$