Relative Velocity/Two-dimensional motion

  • Thread starter Thread starter kingofretards
  • Start date Start date
  • Tags Tags
    Motion Relative
Click For Summary

Homework Help Overview

The discussion revolves around a problem involving the motion of a Chinook salmon as it attempts to jump over a waterfall that is 1.50 m high, starting from a distance of 1.18 m from the base. Participants are tasked with finding the x- and y-components of the initial velocity needed for the salmon to reach the ledge at the top of its trajectory.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need to find the initial velocity components and question how to determine time given the distance and height. Some suggest using trigonometric relationships based on the triangle formed by the height and distance, while others express uncertainty about which formulas to apply.

Discussion Status

There is an ongoing exploration of the relevant equations and concepts, with some participants suggesting that the problem can be approached by considering the forces acting on the salmon. Others note the importance of understanding the motion in both the x and y directions, particularly the effects of gravity on the y-component of velocity.

Contextual Notes

Participants are working under the constraints of needing to find the initial velocity components without explicit guidance on the formulas or methods to use, leading to varied interpretations of the problem setup.

kingofretards
Messages
7
Reaction score
0

Homework Statement


Suppose a Chinook salmon needs to jump a waterfall that is 1.50 m high. If the fish starts from a distance 1.18 m from the base of the ledge over which the waterfall flows, find the x- and y-components of the initial velocity the salmon would need to just reach the ledge at the top of its trajectory.


Homework Equations


x-component
vx=v0*x + ax*t
deltax=v0x*t+.5ax*t^2
vx^2=v0x^2+2*ax*deltax

also y-components, replace x components above

v= (vx^2+vy^2)^.5 (pythagorean theorem)

theta = tan^-1 (vy/vx)

I don't know, since they're looking for the v0x and v0y, or the x- and y- components of velocity.

The Attempt at a Solution


Okay, I only managed to make a diagram, but otherwise, no luck.
 
Physics news on Phys.org
I think they are looking for the added vector of the v0x and the v0y.

You are given the angle from the right triangle (with legs 1.5 and 1.18). You know you need to traverse at least 1.18 m before you reach 1.5 m in height (this can provide a time). And finally, you know the acceleration of gravity. The rest is grunt work.
 
Yes, but I don't know which formulas to use. I just don't know how to find time. I have the formulas ready, but if all i have is the distance, but how can I find time, if I'm looking for velocity?
 
For your problem there is only one formula, F = ma ?

F_x = ma_x = 0 --> v_x = v_ox --> x = x_o + v_ox*t
F_y = ma_y = -mg -->

v_y = v_oy - g*t --> y = y_o + v_oy*t - g*t^2/2
 
Spinnor said:
For your problem there is only one formula, F = ma ?

F_x = ma_x = 0 --> v_x = v_ox --> x = x_o + v_ox*t
F_y = ma_y = -mg -->

v_y = v_oy - g*t --> y = y_o + v_oy*t - g*t^2/2

I don't think force has anything to do with it.
 
kingofretards said:
I don't think force has anything to do with it.

There is no force in the x direction so when the fish "leaps" it will move in the x direction with constant velocity. In the y direction the fish moves with some initial velocity in the y direction but this velocity changes with time as the force of gravity acts on the fish.

Parabolic motion, assume the velocity of the fish in the y direction goes to zero when the fish reaches the ledge.
 
The time can be found with the acceleration of gravity and v0y as the initial velocity in the y direction.
 

Similar threads

Replies
11
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 29 ·
Replies
29
Views
10K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
2
Views
4K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 13 ·
Replies
13
Views
2K
Replies
9
Views
3K