Question

a stone is thrown vertically upward with velocity V from height H from the ground. find the ratio of average velocities of stone from height H to top and from top to ground.

Answer 1

Answer:

Explanation:

Answer 2

Given:

A stone is thrown vertically upwards with a velocity $v$ from a height $H$ above the ground.

To Find:

The ratio of the average velocities of the stone from height $H$ to the top and top to ground.

Solution:

We know that the stone moving under the influence of gravity will have a uniformly accelerated motion with an acceleration, $g$ downwards.

In a uniformly accelerated motion, the average velocity between two points is the arithmetic mean of the velocities at those points.

                      $v_{av}=\frac{v_1+v_2}{2}$

When it is moving upwards, the velocity of the stone at the top will be zero.

So, average velocity in the upward motion,

                        $v_{av}=\frac{v}{2}$

When the stone reaches the bottom, its velocity will be

                 $v'=\sqrt{v^2-2gH}$

Average velocity in the downward motion,

              $v'_{av}=\frac{v'}{2}=\frac{1}{2} \sqrt{v^2-2gH}$

Required ratio $=\frac{v_{av}}{v'_{av}}$

                        $=\frac{v}{\sqrt{v^2-2gH}}$

Hence, the ratio of the average velocities of the stone from height $H$ to the top and top to ground $=\frac{v}{\sqrt{v^2-2gH}}$.