Question

Tarzan, in one tree, sights Jane in another tree. He grabs the end of a vine with length 20 m that makes an angle of 45 with the vertical, steps off his tree limb, and swings down and then up to Jane's open arms. When he arrives, his vine makes an angle of 30 with the vertical. Calculate Tarzan's speed just before he reaches Jane. You can ignore air resistance and the mass of the vine.

Answer 1

This is a new twist at Physics.

Tarzan arrives at Jane with a velocity of 7.9 M/sec.
X______________________
..................../|\
................../..|..\
.....(20M)../.....|30\
............/...45.|......\.(20M)
......../............|........\
.a.O________.|..........\
.....................|____b..O----> Vf = 7.9 M/sec


Tarzan comes from point (a) at a distance of

X==>a = 20 cos 45 = 20 (.707) = 14.14 M

Then he arrives at point (b) at a distance of

X==>b = 20 cos 30 = 20 (.866) = 17.32 M

A difference of (17.32 - 14.14 = 3.18) M, and relative to point (b) Tarzan has a PE at (a)

PE at (a) = mgH
................= mg(3.18)

This PE gives him KE when he arrives at Jane

KE = 1/2 mVf^2

KE = PE

1/2 mVf^2 = mg(3.18)..<===m cancels out

Vf^2 = g (3.18) 2
........= (9.81)(3.18) 2

Vf = 7.90 M/sec

Answer 2

This can be done with the conservation of energy

What we can find first are how high Tarzan is above his lowest point at both ends of his swing.

The angle of the vine makes a right triangle with a vertical. The height will the length of the vine minus the distance upward

So, (20)(cos 45) = 14.14

20 - 14.14 = 5.86 m high to start

(20)(cos 30) = 17.32 m

20 - 17.32 = 2.68 m high at the end

Then, PE at the start = PE + KE at the end

mgh₁ = .5mv² + mgh₂   (m cancels)

(9.8)(5.86) = (.5)(v²) + (9.8)(2.68)

v = 7.89 m/s