Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Projectile motion of a salmon jump

  1. Oct 1, 2003 #1
    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.
  2. jcsd
  3. Oct 1, 2003 #2


    User Avatar
    Science Advisor

    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.
  4. Oct 2, 2003 #3


    User Avatar
    Science Advisor

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook