Projectile motion of a salmon jump

In summary, the conversation discusses a problem involving a salmon jumping over a waterfall and the minimum speed it must have to continue upstream. The solution involves using vector addition and understanding how objects move under constant gravity with no air resistance. The final answer is 6.2 m/s and it is important to think through the problem before attempting to solve it.
  • #1
Gunk
I am having problems with this section. Any help is appreciated.

Salmon often jump water falls to reach their breeding ground. Starting 2 m from a waterfall .55 m in height, at what minimum speed must a salmon jumping at an angle of 32 degrees leave the water to continue upstream?

The answer is 6.2 m/s. The problem is I really don't know where to begin. I figured that I would do some vector addition with 2(sin32) but that didn't lead me anywhere. Thanks for any help.
 
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  • #2
Don't just start putting numbers together without thinking it out.

2 sin(32) would give you a distance (actually, it's 2 cos(32) you would want for height) but that's assuming a straight line jump which just doesn't happen. If you are expected to do problems like this then you should know that objects moving under constant gravity with no air resistance move in parabolas.

This is a fairly standard "acceleration vector, velocity vector, position vector" problem.

You know, since this is a surface of the earth, no air resistance, problem, that the acceleration vector is -9.8 j (j is the vertical unit vector). The initial velocity vector is v0 cos(32)i+ v0 sin(32)j
where v0 is the (unknown) initial speed. (If you don't see how I got that, draw a picture showing the velocity vector at 32 degrees to the water and think "right triangle".)

You should be able to calculate, then, that the "position vector" is
vo cos(32)t i+ ((vo cos(32) t- (9.8/2) t2)j.

Now! When the salmon has moved 2 meters horizontally (to the waterfall), we must have vo cos(32) t= 2. That depends on both v0 and t. At that time, to make it over the waterfall, the salmon must have height at least that of the waterfall: the vertical distance moved must be vo cos(32)- (9.8/2) t2= 5.5.

You can solve those two equation for vo and t.
 
  • #3
Oops. That last equation should be
vo cos(32) t - (9.8/2) t2= 5.5.
You don't really need to solve for t:

From v0 sin(32) t= 2, we have t= 2/(v0 sin(32)).

Put that into the other equation and solve for v0.
 

1. What is projectile motion?

Projectile motion is the movement of an object through the air or space under the influence of gravity. It follows a curved path called a parabola.

2. How does a salmon jump demonstrate projectile motion?

When a salmon jumps out of the water, it is propelled by its powerful tail muscles and follows a curved path through the air before landing back in the water. This motion is an example of projectile motion.

3. What factors affect the projectile motion of a salmon jump?

The factors that affect the projectile motion of a salmon jump include the initial velocity of the jump, the angle at which the salmon jumps, and the force of gravity.

4. How can the trajectory of a salmon jump be calculated?

The trajectory of a salmon jump can be calculated using the equations of motion, taking into account the initial velocity, angle of projection, and the acceleration due to gravity (9.8 m/s^2).

5. Can the projectile motion of a salmon jump be influenced by external factors?

Yes, external factors such as wind resistance, air density, and the surface of the water can affect the trajectory of a salmon jump. These factors can cause the salmon to deviate from its expected parabolic path.

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