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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 degrees 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 degrees with the vertical
Determine whether he gives her a tender embrace or knocks her off her limb by calculating Tarzan's speed just before he reaches Jane
Ignore air resistance and the mass of the vine.

Respuesta :

metchelle
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:

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

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