# Question #e2090

Jan 17, 2017

#### Answer:

$19.8 m {s}^{-} 1$ in the downwards $\downarrow$ direction.

#### Explanation:

The top of the platform is to be taken to be the ground level from which stone is thrown up. Also ignoring air resistance.

Assuming the stone is thrown vertically up $\uparrow$, it will rise and its velocity will decrease due to action of gravity, constant retardation, and will reach a height $h$ where it will stop momentarily as all the initial kinetic energy $\frac{1}{2} m {v}^{2}$ gets converted into its potential energy $m g h$.
Thereafter, it will fall freely under the action of gravity, constant acceleration, and reach ground level, where all the energy will be converted back to its initial kinetic energy.

This implies that the velocity of the stone just reaching the ground will be same as $19.8 m {s}^{-} 1$ in the downwards $\downarrow$ direction.