Tarzan who weighs 688N,swings from a cliff at the end of a vine 18m long . From the top of

The weight of Tarzan = 688 N

To get his mass:

#sf(mg=688" "N)#

#:.##sf(m=688/g=688/9.81=70.13color(white)(x)"kg")#

The change in his potential energy will appear as kinetic energy at the bottom of the swing.

#:.##sf(cancel(m)gDeltah=1/2cancel(m)v^2)#

#:.##sf(v^2=2gDeltah)#

#sf(v=sqrt(2gDeltah)#

#sf(v=sqrt(2xx9.81xx3.2)=sqrt(62.78)=7.923color(white)(x)"m/s")#

The centripetal force is given by:

#sf(F=(mv^2)/(r))#

#:.##sf(F=(70.13xx7.923^2)/(18)=244.6color(white)(x)N)#

Check the answer by @A08. I have neglected to add the weight of Tarzan to the centripetal force.

This is an oddly worded question as the vine should not break.

We know that for an object of mass #m#, moving with a linear velocity #v# in a circle of radius #r#
Centripetal force#=(mv^2)/r# .....(1)

As shown in the figure above, change in potential energy #=mgDeltah#
has been converted in the kinetic energy of Tarzan.
Kinetic energy #1/2mv^2=mgDeltah#
#=>v^2=2gDeltah# ....(2)
We see that velocity is maximum at the bottom of swing.

Also due to movement in a circle there must be a net
force pointing towards the center of the circle to provide the centripetal force. If #T# is tension in the vine at the bottom of swing then
#T-"Weight of Tarzan "W="Centripetal force"# ....(3)

From (1)
#T=mg+(mv^2)/r#
Using (2)
#T=688+(mxx2gDeltah)/r#
Inserting given values
#T=688+(2xx688xx3.2)/18#
#=>T=932.6N#
We see that this tension is less than the breaking capacity of vine given as #950N#

Hence vine does not snap.