Answer to Question #206405 in Physics for Gurang

Question

A stone was thrown vertically upward and it rised 10m above its initial location before it 

started to fall. What was its initial velocity?

Answer 1

According to the energy conservation law, the initial kinetic energy (at the start point) of the stone is equal to its potential energy at the highest point of its trajectory:

$$\dfrac{mv^2}{2} = mgh$$

where $m$ is the mass of the stone, $v$ is its initial velocity, $h = 10m$ is the highest heigth, $g = 9.8N/kg$ is the gravitational acceleration. Expressing $v$, obtain:

$$v = \sqrt{2gh}\\ v = \sqrt{2\cdot 9.8\cdot 10} = 14m/s$$

Answer. 14 m/s.