Question
Tarzan swings on a 33.0 m long vine initially
inclined at an angle of 31.0◦ with the vertical.
The acceleration of gravity if 9.81 m/s
What is his speed at the bottom of the
swing if he
a) starts from rest?
Answer in units of m/s
b) pushes off with a speed of 5.00 m/s?
Answer in units of m/s.
Thanks!
inclined at an angle of 31.0◦ with the vertical.
The acceleration of gravity if 9.81 m/s
What is his speed at the bottom of the
swing if he
a) starts from rest?
Answer in units of m/s
b) pushes off with a speed of 5.00 m/s?
Answer in units of m/s.
Thanks!
Answers
height above bottom call it h
h = 33(1-cos 31)
initial energy = m g h
speed at bottom = sqrt(2gh)
because
(1/2)mv^2 = m g h
for part b
initial energy = m g h +(1/2)m(25)
so at bottom
(1/2) m v^2 = that initial energy
h = 33(1-cos 31)
initial energy = m g h
speed at bottom = sqrt(2gh)
because
(1/2)mv^2 = m g h
for part b
initial energy = m g h +(1/2)m(25)
so at bottom
(1/2) m v^2 = that initial energy
Why you put 12(1-cos 31degrees)
Related Questions
Tarzan swings on a 27.0 m long vine initially inclined at an angle of 35.0° with the vertical.(a) Wh...
Tarzan swings on a 27.0 m long vine initially inclined at an angle of 33.0° with the vertical.
(a)...
Tarzan is often portrayed as swinging through the jungle from vine to vine. Each of Tarzan's vines a...
Tarzan swings on a 30.0 - m - long vine initially inclined at an angle of 37.0 degrees with the vert...