A stone is dropped from the top of an overhead water tank and at the same time another stone is thrown vertically upwards from the ground level with a velocity of 20 ms^-1. If the stones met exactly 2 seconds after their release, determine the height of the overhead tank

To solve this problem, we need to use the equations of motion.

1. Let's consider the stone that is dropped from the top of the tank first. The distance it travels in 2 seconds can be calculated using the equation of motion:

   h1 = ut + 0.5gt^2

   Where: u is the initial velocity = 0 (since the stone is dropped, not thrown) g is the acceleration due to gravity = 9.8 m/s^2 t is the time = 2 seconds

   Substituting these values into the equation, we get:

   h1 = 02 + 0.59.8*(2^2) = 19.6 m

2. Now, let's consider the stone that is thrown upwards. The distance it travels in 2 seconds can be calculated using the equation of motion:

   h2 = ut - 0.5gt^2

   Where: u is the initial velocity = 20 m/s g is the acceleration due to gravity = 9.8 m/s^2 t is the time = 2 seconds

   Substituting these values into the equation, we get:

   h2 = 202 - 0.59.8*(2^2) = 20.4 m

3. The total height of the tank is the sum of the distances travelled by the two stones, so:

   H = h1 + h2 = 19.6 m + 20.4 m = 40 m

Therefore, the height of the overhead tank is 40 meters.