Relative Velocity/Two-dimensional motion

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To determine the initial velocity components required for a Chinook salmon to jump a 1.50 m waterfall from a distance of 1.18 m, the discussion emphasizes the need to analyze both x and y motion separately. The x-component of velocity remains constant, while the y-component is affected by gravity, necessitating the use of kinematic equations. A right triangle is formed with the waterfall height and distance, allowing for the calculation of the angle and subsequent determination of time. The key challenge is finding the time of flight, which can be derived from the y-component's equations of motion. Ultimately, understanding the parabolic trajectory and applying the correct formulas will yield the required initial velocity components.
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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.
 
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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.
 
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