Sir solve this

u1: velocity with which the ball starts to fall

v1: velocity with which it hits the ground

u2: velocity after collision with ground

v2: velocity at highest point after it hits the ground = 0

 

now 

v₁² = u₁² + 2ah = u₁² + 2(10)(10) = u₁² + 200  {a = 10 m/s2, moving due to gravity}


Ball loses half of the energy in collision.

thus, (mu₂²)/2= (mv₁²)/4

         u₂² = v₁^2/2 = (u₁² + 200)/2

 

And, v₂² = u₂² + 2ah = [(u₁² + 200)/2] - 2(10)(10)         {a = -10 m/s2, moving against gravity}

 

also v₂ = 0

Gives u₁ = 10√2 m/s