Question

Tarzan, in one tree, sights Jane in another tree

Answer 1

Taking the distance of Tarzan from the ground before and after he makes the swing:

Ho (initial height) = L(1 - cos45) = 20 (1 - 0.707) = 5.86 meters
Hf (final height) = L(1 - cos30) = 20 (1 - 0.866) = 2.68 meters

Difference in height = 5.86 - 2.68 = 3.18 meters

PE = KE
mgh = (1/2)mv^2

Solving for v:
v = sqrt (2*g*h)
v = sqrt (2*9.8*3.18)
v = 7.89 m/s

With Tarzan going that fast, it is likely that he will knock Jane off.

Answer 2

Answer:

Tarzan's speed is 7.91m/s.

Tarzan will knock off Janefrom the tree.

Explanation:

Initial vertical height = L-LcosA = L(1-cos45)

Initial vertical height = 20× 0.29= 5.8m

Final vertical height = L(1-cos30°)= 20× 0.13

Final vertical height = 2.6m

Loss in vertical and = 5.8- 2.6=3.2m

Loss in Potential energy = Gain in kinetic energy

Mgh=1/2(mv^2)

V= sqrt( 2× 9.8 × 3.2)

V= 7.9m/s