A body of mass `m` is dropped from a height of `h`. Simultaneously another body of mass `2m` is thrown up vertically with such a velocity `v` that they collide at height `h/2`. If the collision is perfectly inelastic, the velocity of combined mass at the time of collision with the ground will be

To solve this problem, we need to use the principles of conservation of momentum and energy.

Step 1: Calculate the velocity of the first body when it reaches height h/2.

We can use the equation of motion v² = u² + 2gs, where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and s is the distance. Since the body is dropped, u = 0. So, the equation becomes v² = 2g(h/2). Solving for v, we get v = sqrt(g*h).

Step 2: Calculate the velocity of the second body when it reaches height h/2.

We can use the same equation of motion. The initial velocity is v, the final velocity is 0 (at the maximum height), and the distance is h - h/2 = h/2. So, the equation becomes 0 = v² - 2g(h/2). Solving for v, we get v = sqrt(g*h).

Step 3: Calculate the velocity of the combined mass after the collision.

The total momentum before the collision is msqrt(gh) + 2m*(-sqrt(gh)) = 0 (since the two bodies are moving in opposite directions). The total momentum after the collision is (m + 2m)v' = 3mv', where v' is the velocity of the combined mass. Since momentum is conserved, we have 3mv' = 0, which implies v' = 0.

Step 4: Calculate the velocity of the combined mass when it hits the ground.

The combined mass falls from a height of h/2 with initial velocity 0. Using the equation of motion v² = u² + 2gs, we get v² = 2g(h/2). Solving for v, we get v = sqrt(g*h).

So, the velocity of the combined mass at the time of collision with the ground is sqrt(g*h).