Projectile Motion of a Salmon: Finding Minimum Speed for a Waterfall Jump

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Homework Help Overview

The problem involves a salmon attempting to jump a waterfall, requiring the determination of the minimum speed needed to reach a height of 0.55 meters while starting 2 meters away from the waterfall at an angle of 32 degrees. The subject area is projectile motion.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss using the range equation and other kinematic equations to find the initial velocity. There are questions about the relationship between the y and x components of velocity and how to apply the equations correctly.

Discussion Status

Some participants suggest different methods for calculating the initial velocity, including the range formula and kinematic equations. There is confusion regarding the results from different approaches, and some guidance has been offered to focus on specific equations. However, there is no explicit consensus on which method is preferred.

Contextual Notes

Participants express uncertainty about the application of the range formula and its relevance to the problem. There is also mention of differing results when applying various methods, leading to confusion about which approach aligns with the expectations of the instructor.

princesspriya
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Homework Statement


salmon often jump waterfalls to reach their breeding grounds. One salmon starts 2m from a waterfall that is .55m tall and jumps at an angle of 32degrees. what must be the salmon's minimum speed to reach the waterfall?


Homework Equations





The Attempt at a Solution


so the distance in y is .55 and distance in x is 2 and the degree is 32 and trying to find Vi. but i m lost like how would i be able to solve for Vi if i do not have any of the velocity in x or y direction.
 
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you can try using the range equation (meaning total distance in the x direction) since you have the distance X and the angle the formula is R=(V^2/g)sin2a where a=the angle

Or since you may not have learned this you can try solving for the velocity using V^2=V^2 + 2a(y-y). The trick here is realizing that your velocity in the y direction will be 0 at the top of his arc at .55m so then you can solve for the initial.
 
for the second equation you provided, is it the same as Vf^2=Vi^2(sinX)^2+2ay
 
I have never had to use a range formula in projectile motion. It is a nice thought, but in doing so, many miss out on the physical meaning behind its derivation. mer584 had the right idea on his/her second thought.

If it is the minimum initial velocity to reach the waterfall, what must the final velocity in the y direction be?
 
princesspriya said:
for the second equation you provided, is it the same as Vf^2=Vi^2(sinX)^2+2ay

well...it is Vf^2= Vi^2 + 2a(Yf-Yi) or delta y.
It will give you the y component of the velocity when you solve for Vi
multiplying it by sin(a) where a=angle will provide you with the the y component IF you were starting with the total velocity. BUT since you are starting just with the y component you will not need sin just yet.

you will need it, however, to find the total at the end Vy= V(sin a)

So to reiterate you would use the formula I mentioned to solve for Vy then that last one above to solve for V

combining that in one step is confusing to me, but perhaps you can see how it connects now to the formula you gave
 
idk because wen i plug numbers in i get a difference answer using different methods(the one you provided and the range formula) for the range formula i got 4.67 and the other one i got 6.19 so i am confused on which one to use?
 
ignore the range formula, the other method is definitely what your prof will want

but for reference the range formula will give you the total initial velocity whereas the large formula we have been discussing will give you the y component and then allow you to solve for the total initial. But honestly...ignore it, sorry.
 

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