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Grade: 11
        
A man of mass M having a bag of mass m slips from a roof of tall building of height h and starts falling vertically . when at  a height of h m from the ground he notices that there is a pretty hard ground below him but there is a pond at horizontal distance x from the line of fall . in order to save himself he throws the bag horizontally in the direction opposite to tjc pond . calculate the min. Horizontal velocity imparted to the bag to avoid the hard ground where will the bag land.
 
2 years ago

Answers : (1)

Arun
24739 Points
							
Dear Mahak
 
 

4875-2266_5850_untitled.JPG

let u,v be horizontal velocities of man and bag

balancing moments gives : Mu = mv

and h=1/2.g.t2   , x = ut

from all 3 eqns , we have hz. velocity of bag is  v = Mx /[m.(2h/g)1/2 ]

now tha bag will land at , y = vt

on solving all eqns we have

bag lands at : y = Mx/m

 

 

Regards

Arun (askIITians forum expert)

2 years ago
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