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
Answer:
a) [tex]v_{f} \approx 0.328\,\frac{m}{s}[/tex], b) [tex]v_{f} \approx 6.009\,\frac{m}{s}[/tex]
Explanation:
Let consider that bottom has a height of zero. The motion of Tarzan can be modelled after the Principle of Energy Conservation:
[tex]U_{g,1} + K_{1} = U_{g,2} + K_{2}[/tex]
The final speed is:
[tex]K_{2} = U_{g,1} - U_{g,2} + K_{1}[/tex]
[tex]\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}[/tex]
[tex]v_{f}^{2} = 2 \cdot g \cdot L \cdot (\cos \theta_{2} - \cos \theta_{1}) + v_{o}^{2}[/tex]
[tex]v_{f} = \sqrt{v_{o}^{2}+2\cdot g \cdot L \cdot (\cos \theta_{2}-\cos \theta_{1})}[/tex]
a) The final speed is:
[tex]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})}[/tex]
[tex]v_{f} \approx 0.328\,\frac{m}{s}[/tex]
b) The final speed is:
[tex]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})}[/tex]
[tex]v_{f} \approx 6.009\,\frac{m}{s}[/tex]
