Given

#L->"Length of the rope"=25m#

#M->"Mass of the man climbing"=72kg#

#T->"Tension the rope can withstand"=2.9kN#

Let the **sagging be x m** when the man is at the **mid point** of the rope and the rope then subtends angle #theta# with the horizontal.

The two vertical components of tension will balance the weight of the man.

So

#2Tsintheta=Mxxg#

#sintheta=(Mg)/(2T).....(1)#

For #theta# being small #sintheta~~tantheta=x/(L/2)=(2x)/L#

Now **imposing approximation** the relation (1) becomes

#(2x)/L=(Mxxg)/(2T)#

#=>x=(MxxgxxL)/(4T)#

#=>x=(72xx9.8xx25)/(4xx2900)~~1.52m#

**Without approximation** the realation (1) can be written as

#x/sqrt(x^2+(L/2)^2)=(Mxxg)/(2T)#

#=>(x^2+(12.5)^2)/x^2=((2xx2900)/(72xx9.8))^2=67.57#

#=>1+156.25/x^2=67.57#

#=>156.25/x^2=67.57-1=66.57#

#=>x^2=156.25/66.57#

#=>x=sqrt(156.25/66.57)~~1.53m#

**Another way** of calculation

From (1)

#theta=sin^-1((Mxxg)/(2T))=sin^-1((72*9.8)/(2*2900))#

#theta=6.99^@#

So

#(2x)/L=tantheta=tan6.99#

#x=(25xxtan6.99)/2=1.53m#