What is his speed at the bottom of the swing if he does the following?

Question

Tarzan swings on a 25.0 m long vine initially inclined at an angle of 41.0° with the vertical. What is his speed at the bottom of the swing if he does the following?

(a) He starts from rest.
10.9632 m/s
(b) He pushes off with a speed of 4.00 m/s.
_____ m/s

Understanding the relationship between potential energy, kinetic energy, the conservation of energy and work.

Answer 1

Using conservation of energy the kinetic energy at the bottom must equal the initial energy, that is the sum of potential=mgh and kinetic(for part b),
Find h by assuming the path is a circle with radius 25 meters.

Answer 2

a) Potential Energy (PE) + Kinetic Energy (KE) at the start= PE+KE at the bottom.

At start: PE=mgh
h= (25 - 25cos41)

PE=mg(25-25cos41)
KE=1/2mv^2
v=0 because he starts from rest.
KE=0

At bottom

PE =mgh
h=0
PE=0

Ke=1/2mv^2


mg(25-25cos41)+0=0 + 1/2mv^2
m cancels out.

g(25-25cos41)=1/2v^2
g=9.8m/s^2

You have one unknown: v. You can solve for v, can't you?

b)At start, same PE, but now KE=1/2m*4^2

At bottom same PE=0; same KE=1/2mv^2

mg(25-25cos41)+1/2m*4^2=0 +1/2mv^2

m cancels out.

g(25-25cos41)+8=1/2v^2
9=9.8m/s^2

You can solve for v, can't you?