Man swing from on a rope (velocity)

AI Thread Summary
A man swings from a 15 m high cliff on a 12 m rope, releasing at a 30° angle to the horizontal. The discussion focuses on calculating his velocity at the release point and the distance from the landing spot to the cliff base. The velocity is derived using energy conservation principles, leading to a calculated speed of 10.85 m/s. The angle of velocity is clarified to be 50° due to the geometry of the swing. The conversation emphasizes the importance of understanding the relationship between the swing's radius and the velocity's direction.
AdsBerger
Messages
13
Reaction score
0

Homework Statement


A man standing on a 15 m high cliff tries to swing on a 12 m rope and let's go at 30° of the horizontal (travelling 150°). He let's go and travels through the air, to then land on the ground.

What is his distance from his landing spot to the base of the cliff?


Homework Equations





The Attempt at a Solution



I am able to figure out the equation to find the height from the ground of which he let's go.
Not able to figure out the velocity when he let's go.
There is no mass for him, so those equations are irrelevant. If he is starting at a horizontal angle, then I am unsure of what equations to use to find his velocity.
 

Attachments

  • Physics problem.png
    Physics problem.png
    10.7 KB · Views: 1,551
Physics news on Phys.org
The man has mass m. The actual value of m is irrelevant, you will see. What equations you are considering irrelevant?

The angle of the velocity is not 30°with respect to the horizontal. The man swings along a circle and the velocity is tangent to the circle, so perpendicular to the radius.ehild
 
Last edited by a moderator:
I am thinking i will have to use the equation v = √{2gL[1-cos(a)]}
v = √{2(9.8)(12) [1-cos(a)]}
What would the angle be though?
 
To find the velocity ##V## one may use the energy conservation law:
$$
m \frac {v^2} {2} = m g Δh.
$$

So mass ##m## will cancel and the difference of heights ##Δh## you may find out using geometry laws.
 
Consider:
1. can you use energy conservation (k.e. + p.e. = constant) or does the rope exert a force on the man to add to or subtract from his total energy?

2. Once you know v at the release point you are all set. Right?
 
AdsBerger said:
I am thinking i will have to use the equation v = √{2gL[1-cos(a)]}
v = √{2(9.8)(12) [1-cos(a)]}
What would the angle be though?

Where is that equation from?

As for the angle of velocity look at the picture. The velocity is normal to the radius. What is the angle beta?

ehild
 

Attachments

  • manrope.JPG
    manrope.JPG
    7.3 KB · Views: 608
The angle beta would be 50. i calculated his velocity to be 10.85 m/s
 
AdsBerger said:
The angle beta would be 50. i calculated his velocity to be 10.85 m/s

Why 50? ? It is a right-angled triangle (the yellow one) and one angle is 30°, beta is the other angle...

ehild
 
Back
Top