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

Answer:

a) $v_{f} \approx 0.328\,\frac{m}{s}$, b) $v_{f} \approx 6.009\,\frac{m}{s}$

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:

$K_{2} = U_{g,1} - U_{g,2} + K_{1}$

$\frac{1}{2}\cdot m \cdot v_{f}^{2} = m\cdot g \cdot L\cdot (\cos \theta_{2}-\cos \theta_{1}) + \frac{1}{2}\cdot m \cdot v_{o}^{2}$

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

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

a) The final speed is:

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

$v_{f} \approx 0.328\,\frac{m}{s}$

b) The final speed is:

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

$v_{f} \approx 6.009\,\frac{m}{s}$

Answer 2

Answer:

(A) Vf = 13.8 m/s

(B)  Vf = 15.1 m/s      

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