Question

From top of a very high tower, two stones are thrown simultaneously vertically up and vertically down respectively, each with same speed of 10 m/s. After few seconds what is magnitude of their relative velocity (in m/s) when stones are still in air?​

Answer 1

Explanation:

As the stones travel in their respective direction the stone going downwards gains more speed due to gravity increasing its force and the stone going upward loses speed due to gravity decreasing it's force. As they. have been thrown at a uniform speed from the beginning they should also slow or speed up at the same rate.

Answer 2

Given: From top of a very high tower, two stones are thrown simultaneously vertically up and vertically down respectively, each with same speed of 10 m/s.

To find: After few seconds the magnitude of their relative velocity (in m/s) when stones are still in air.

Solution:

The velocity of an object in uniform motion is calculated using the following formula.

$v = u + at$

Here, v is the final velocity, u is the initial velocity, a is the acceleration acting on the stones and t is the time in seconds.

Let the few seconds of time be t seconds. The acceleration acting on both the stones is the acceleration due to gravity and the acceleration against gravity. The velocities of the stone thrown upwards and the stone thrown downwards is represented by the following equations.

$v_{up} = 10ms^{-1} + (-10 ms^{-2})t$

      $= 10 - 10 t$

$v_{down} = 10ms^{-1} + (10 ms^{-2})t$

         $= 10 + 10t$

Now, the relative velocity between the velocites of the two stones is

$v_{relative} = v_{down} - v_{up}$

             $= 10 + 10t - (10 - 10t)$

             $= 20t$

Therefore, after few seconds the magnitude of their relative velocity (in m/s) when stones are still in air is 20t ms⁻¹.