A man of mass m having a bag ofmas m slips from the roof of atall building of height H and starts falling vertically figure. When at a height h from the ground, he notices that the ground below him is pretty hard but there is a pond at a horizontal distnce x from the line of fall. In order to save himself he throws the bag horizontally (with respect to himself) in the direction opposite to the pond. Calculate the minimum horizontal velocity imparted to the bag so that the man lands in the water. If the man just suceeds toavoid the hard ground, where will the bag land?

Explanation

Mass of man =M,
Initial velocity =O Mass of bag = m
Let the man throws the bag towards left with a velocity v.
So there is no exterN/Al force in the horizontal direction

The momentum will be conserved.
Let the goes right with a velocity
`mv=MV`
rarr V=((mv)/M)`
`rarr v=((MV)/m)`............i
Let the total time he will take to reach grond
`=sqrt((2H)/g)=t_1`
Let the total time he will take to reach the height h
`=t_2=sqrt((2(H-h))/g)`
Then the time of his flying
`=t_1-t_2`
`=((2h)/g-)sqrt((2(H-h))/g)`
`=sqrt(2/g)(sqrtH-sqrt(H-h)` ltbr. with in this time he reaches the ground in the pond covering a horizontal distance
`rarr x=Vxxt `
`rarr V=x/t`
`v=M/m x/t`
`=M/mxxxx sqrtg/(sqrt(2)(sqrtH-sqrt(H-h))x`
As there is no exterN/Al force in horizontal direction the x-corrdiN/Ate of CM will remain at that position.
`rarr 0=(Mxx(x)+mxxx_1)/(M+m)`
`rarr x_1=M/mx`
`:.` The bag will reach the bottom at a distance `(M/m)x` towards left of the line it falls.