Question Number 105037 by Dwaipayan Shikari last updated on 25/Jul/20

A heavy man , slips  from a roof of a high building of  height ′H′. When he was at a height of ′h′  he realised that the  ground beneath him was pretty hard. To avoid injury, he thrown  away his suitcase of mass ′m′. After doing this,he prevented his  falling on the hard ground and fell on a pond. The mass of the  man was ′M′. The distance of the pond from the hard ground was   ′x′.And he just avoided from falling on the hard ground.    Find the velocity required to throw away the suitcase as the man  don′t fall on the hard ground.(There was no horizontal velocity of the  suitcase−man system)

Answered by mr W last updated on 25/Jul/20

time for the fall from height h to  ground:  t=(√((2H)/g))−(√((2(H−h))/g))=(√((2H)/g))(1−(√(1−(h/H))))  horizontal velocity required for the  man, in order not to fall on the hard  ground:  u_M =(x/t)=(x/((1−(√(1−(h/H))))(√((2H)/g))))  say the horizontal velocity of  suitcase is u_S ,  mu_S =Mu_M   ⇒u_S =(M/m)u_M   required velocity to throw away the  suitcase is  v=u_S +u_M =(1+(M/m))u_M   ⇒v=(1+(M/m))(x/((1−(√(1−(h/H))))(√((2H)/g))))

Commented byDwaipayan Shikari last updated on 25/Jul/20

Great sir!