PLEASE HELP WITH THE ABOVE PROBLEM:) USE THE SAME EXACT MEASUREMENTS Consider an

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PLEASE HELP WITH THE ABOVE PROBLEM:) USE THE SAME EXACT MEASUREMENTS

Consider an ideal pendulum of length 15.0m with an initial angle of 66.0 degrees from the vertical. A mass of 90.0kg (Tarzan) is attached to the end of the rope (vine), initially at rest. Determine:

a) The SPEED of Tarzan at the bottom of his swing

b) The TENSION T in the vine at the above position.

c) The VELOCITY of Tarzan when he releases the vine at an angle of 25.0 degrees on the other side of the vertical.

d)After releasing the vine, what is the maximum height, measured from the bottom of the swing, will he reach during his trajectory?

e) If a river with deep banks is located 14.0m away from his release point, does he land in the river? If so, by how much did he clear the nearest bank? Assume that the ground is 16.0m below the attachment point of the vine.

Explanation / Answer

15 - (cos 66 x 15) = 8.899 metres above starting point.
PE = (mgh) = 90 x 9.8 x 8.899 = 7,848.92 Joules.
This becomes KE at the bottom of the arc.

a) V = sqrt.(2KE/m), = 13.2m/sec.

b) T = (mv^2/r) = 1,045.44N.
15 - (cos 25 x 15) = 1.4 metres.
PE = (mgh) = 90 x 9.8 x 1.4, = 1,234.8 Joules.
(7,848.92 - 1,234.8) = 6,614.12 Joules.

c) V at release = sqrt.(2KE/m) = 12.12m/sec., at 25 degrees above horizontal.
Horizontal component of trajectory = (cos 25) x 12.12, = 10.98m/sec.
Initial vertical component = (sin 25) x 12.2, = 5.12m/sec.
Height gain after release = (v^2/2g) = (5.12^2/19.6) = 1.34 metres.

d) (1.34 + 1.4) = 2.74 metres above bottom of swing.
The ground is (16 - 15) = 1 metre below the bottom of the swing, so height above ground at top of trajectory = (2.74 + 1) = 3.74 metres.
Time to drop 3.74 metres = sqrt.(2h/g) = sqrt.(3.74 x 2)/9.8, = 0.874 sec.
Time from release to top of trajectory = (v/g) = 5.12/9.8, = 0.522 sec.
Total time = (0.522 + 0.874) = 1.396 secs.
Horizontal distance covered = (1.396 x 10.98) = 15.33 metres.

e) He's in the river. Now, not sure if you need vertical or horizontal clearances, but horizontal clearance is 1.33 metres.
Time to travel 14m. horizontally = (14/ 10.98) = 1.275 secs.
Time from top of trajectory to near bank = (1.275 - 0.522) = 0.753 secs.
In that time, he would fall 1/2 (t^2 x g) = 2.78 metres.
(3.74 - 2.78) = vertical clearance of bank of 0.96 metre.

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