Question

Tarzan swings on a 35.0 m long vine initially inclined at an angle of 44.0◦ with the vertical. The acceleration of gravity if 9.81 m/s2.

What is his speed at the bottom of the swing if he
a) starts from rest?
b) pushes off with a speed of 6.00 m/s?

Answer 1

a)
`v_(f) \approx 0.328\,(m)/(s)`, b)
`v_(f) \approx 6.009\,(m)/(s)`

Step-by-step explanation:

Let consider that bottom has a height of zero. The motion of Tarzan can be modelled after the Principle of Energy Conservation:

`U_(g,1) + K_(1) = U_(g,2) + K_(2)`

The final speed is:

`K_(2) = U_(g,1) - U_(g,2) + K_(1)`

`(1)/(2)\cdot m \cdot v_(f)^(2) = m\cdot g \cdot L\cdot (\cos \theta_(2)-\cos \theta_(1)) + (1)/(2)\cdot m \cdot v_(o)^(2)`

`v_(f)^(2) = 2 \cdot g \cdot L \cdot (\cos \theta_(2) - \cos \theta_(1)) + v_(o)^(2)`

`v_(f) = \sqrt{v_(o)^(2)+2\cdot g \cdot L \cdot (\cos \theta_(2)-\cos \theta_(1))}`

a) The final speed is:

`v_(f) = \sqrt{(0\,(m)/(s) )^(2)+2\cdot (9.807\,(m)/(s^(2)) )\cdot (35\,m)\cdot (\cos 0^(\textdegree)-\cos 44^(\textdegree))}`

`v_(f) \approx 0.328\,(m)/(s)`

b) The final speed is:

`v_(f) = \sqrt{(6\,(m)/(s) )^(2)+2\cdot (9.807\,(m)/(s^(2)) )\cdot (35\,m)\cdot (\cos 0^(\textdegree)-\cos 44^(\textdegree))}`

`v_(f) \approx 6.009\,(m)/(s)`

Answer 2

(A) Vf = 13.8 m/s

(B) Vf = 15.1 m/s

Step-by-step explanation:

length of rope (L) = 35 m

angle to the vertical = 44 degrees

acceleration due to gravity (g) = 9.8 m/s^{2}

(A) from conservation of energy

final kinetic energy + final potential energy = initial kinetic energy + initial potential energy

0.5m(Vf)^{2} + mg(Hf) = 0.5m(Vi)^{2} + mg(Hi)

where

m = mass

Hi = initial height = 35 cos 44 = 25.17

Hf = final height = length of vine = 35 m

Vi = initial velocity = 0 since he starts from rest

Vf = final velocity

the equation now becomes

0.5m(Vf)^{2} + mg(Hf) = mg(Hi)

0.5m(Vf)^{2} = mg (Hi - Hf)

0.5(Vf)^{2} = g (Hi - Hf)

0.5(Vf)^{2} = 9.8 x (25.17 - 35)

0.5(Vf)^{2} = - 96.3 (the negative sign tells us the direction of motion is downwards)

Vf = 13.8 m/s

(B) when the initial velocity is 6 m/s the equation remains

0.5m(Vf)^{2} + mg(Hf) = 0.5m(Vi)^{2} + mg(Hi)

m(0.5(Vf)^{2} + g(Hf)) = m(0.5(Vi)^{2} + g(Hi))

0.5(Vf)^{2} + g(Hf) = 0.5(Vi)^{2} + g(Hi)

0.5(Vf)^{2} = 0.5(Vi)^{2} + g(Hi) - g(Hf)

0.5(Vf)^{2} = 0.5(6)^{2} + (9.8 x (25.17 - 35))

0.5(Vf)^{2} = -114.3 ( just as above, the negative sign tells us the direction of motion is downwards)

Vf = 15.1 m/s