Sir solve this

Asked by spuneet23 | 9th Sep, 2010, 10:50: PM

Expert Answer:

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) = u1+ 200  {a = 10 m/s2, moving due to gravity}


Ball loses half of the energy in collision.
thus, (mu22)/2= (mv12)/4
         u22 = v12/2 = (u1+ 200)/2
 
And, v22 = u22 + 2ah = [(u1+ 200)/2] - 2(10)(10)         {a = -10 m/s2, moving against gravity}
 
also v2 = 0
Gives u1 = 10√2 m/s

Answered by  | 10th Sep, 2010, 02:25: PM

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