Question

Salmon often jump waterfalls to reach their breeding grounds. Starting 1.00m from a waterfall 0.55 m height, what minimum speed must a salmon jumping at 32.0 degrees leave the water to make it over the waterfall?
I know what the angle is and acceleration is gravity and my teacher gave me the answer that the initial vertical velocity is 6.2 m/s but I was wondering how to get there. There are also only a few equations he likes us to use that he calls the basic 7.... If somebody could explain this to me I would be oh so grateful.

Answer 1

hey you can use speed formula ,you know it know

Answer 2

Wait I don't think that works... we don't necessarily want the max height to be 0.55m. It just needs to be 0.55m after the salmon travels 1m horizontally.

Answer 3

initial vertical speed is vsin32. horizontal is vcos32. We'll need both.

Use this (there's no acceleration in the horizontal direction)
\[\Large \Delta x = v_x t \]and this \[\Large \Delta y = v_y t + 0.5 a t^2\]
Delta x is the 1m. Delta y is the 0.55m. vx and vy are the initial horiz and vert velocities
a is -9.8, t is the time when the salmon reaches the point where it crosses the waterfall.
so..
\[\Large 1= (v \cos 32) t \]^solve this for t...then plug it into this:
\[\Large 0.55= (v \sin 32) t + 0.5 (-9.8) t^2\]
which you can then solve for v (you'll need to do some algebra to simplify).

Answer 4

Should look like this after plugging in. Which looks complex, but it isn't really... simplify it a bit and it's not too menacing.\[\Large 0.55= (v \sin 32) \frac{ 1 }{ v \cos 32 } + 0.5 (-9.8) \left( \frac{ 1 }{ v \cos 32 } \right)^2\]

Answer 5

@bunnyxd2 hopefully you follow some of this. It's not an easy problem.

Answer 6

Btw i think your teacher might be wrong with this "my teacher gave me the answer that the initial vertical velocity is 6.2 m/s" - this was my initial answer, but it assumes the max height is 0.55 m, which isn't necessarily the case.

Answer 7

Haha thanks for saying that. :) I just thought I was really dumb.... So what do I do when I have simplified the sine and cosines and gravity? I just don't know what to do with all of the v's? Oh and okay thanks.

Answer 8

lol don't feel dumb for getting stuck on a problem like this! I teach AP physics and this is one of the harder motion problems you'll come across.

simplifying just a little...
\[\Large 0.55= \frac{ v \sin 32 }{ v \cos 32 } - \frac{ 4.9 }{ v^2 \cos^2 32 } \]notice you can cancel/simplify some more before getting v

Answer 9

\[\Large 0.55=\tan 32 - \frac{ 4.9 }{ v^2 \cos^2 32 }\]then
\[\Large 0.55 - \tan 32 = - \frac{ 4.9 }{ v^2 \cos^2 32 }\]remember that all the cos 32 and tan32 are just numbers... you can get them from a calculator.

Answer 10

replacing some of those things with numbers\[\Large -0.07487 = - \frac{ 4.9 }{ v^2 *0.7192} \] you can drop the negatives now and hopefully find v^2, then take square root.

Answer 11

If you multiply both sides by v^2, you should find it's easy to solve.

Answer 12

Okay I feel like you did a REALLY AWESOME job of explaining this but my mind just isn't made for science and physics.... Was the answer you ended up with 9.539?

Answer 13

Yes :)

But if you're using the Giancoli book, this would def be one of the ones labeled with a III (ie meaning it's a challenging one)... it's not an easy or straightforward one.

Answer 14

I'm using the Holt book and I honestly don't see why my teacher is giving these to us. We're only in honors and he will give one example for like 5 sections of homework. It honestly just frustrates me so much....

Answer 15

Yeah that makes it tough :/