What is the speed of tarzan during his swing?

Question

Tarzan swings on a 25.0 m long vine initially inclined at an angle of 34.0° with the vertical.
(a) What is his speed at the bottom of the swing if he starts from rest?
m/s
(b) What is his speed at the bottom of the swing if he pushes off with a speed of 3.00 m/s?
m/s

Answer 1

Use conservation of Mechanical energy
change of gravitation potential energy = change of kinetic energy

change of elevation of the object = 25 - 25Cos 34
= 25(1 -0.829)
= 25* 0.171
= 4.275 m

Δmgh = (1/2)mv²v' ²
v = √[2g(Δh)]
v = √[2*9.8* 4.275]
v = √83.79
v = 9.15 m/s


so inItial (1/2)mv² + mgh = final (1/2)mv' ²
v²/2 + gh = v' ² / 2
9/2 + 9.8 * 4.275 = v' ² / 2
v' ² / 2 = 4.5 + 41.895
v' ² = 92.79
v' = 9.63 m/s

Answer 2

a) Start by using trigonometry to find Tarzan's initial height above the bottom of his swing. With vine length L and angle θ, both given, the height h will be L - L*cosθ. You can then use conservation of energy, where Tarzan's initial gravitational potential energy (no kinetic energy at rest) will equal his final kinetic energy (no GPE at bottom point defined as zero). That is, 0.5*mv^2 = mgh ==> 0.5*v^2 = gh ==> v = sqrt(2gh), where g is known (9.8 m/s^2), v is the unknown velocity, and h was calculated.

b) It's very similar, except that now Tarzan's final KE are equal to his initial GPE and KE. With the unknown velocity still represented as v but now adding the KE due to the initial velocity of 3 m/s, we have 0.5*mv^2 = mgh + 0.5*m*3^2 ==> 0.5*v^2 = gh + 4.5 ==> v = sqrt(2gh + 9), where g and h are still known.

Answer 3

a) Tarzan swings part of a segment of a circle of radius 25m and angle 34 degrees.
Arc length = r theta = 25 x 34/180 x pi = 85/18 pi
Downward acceleration = 9.8 ms-2
v^2 = u^2 + 2as
V^2 = 0 + 2 x 9.8 x 85/18pi = 290.77
v = 17.1ms-1

b) v2=3^2 + 2 x 9.8 x 85/18pi
v = 17.31ms-1