A particle of mass m is dropped from a height h above the ground. At the same time another particle of the same mass is thrown vertically upwards from the ground with a speed of $\sqrt{2gh}$ . If they collide head-on completely inelastically, the time taken for the combined mass to reach the ground, in units of  $\sqrt{\frac{h}{g}}$ is :

detailed solution

Correct option is A

Time for collision  t1=relative distancerelative velocity=h2gh   [relative acceleration is zero]After    t1   ,    velocity of A =VA=0−gt1=−gh2       and    velocity of B=VB=2gh−gt1=gh2−12=gh2At the time of collision , apply momentum conservation           P→i=P→f             ⇒mV→A+mV→B=2mV→f    ⇒-gh2   +gh2    =2V→f ⇒ Vf=0 and Height from ground at time of collision ,  h'=h−12gt12=h−h4=3h4  So time taken for the combined mass to reach the ground =   2h'g=2×3h4g=3h2g=32hg