# Finding speed when fish jumps out of water

1. Sep 22, 2008

### sydneyown

1. The problem statement, all variables and given/known data
Salmon, swimming up the Fraser river to their spawning grounds, leap over all sorts of obstacles. The unofficial salmon-altitude record is an amazing 3.44 m height jump. Assuming the fish took off at 45.0o, what was its speed on emerging from the water? Ignore friction.

2. Relevant equations
?

3. The attempt at a solution
Vxo= Vo Cos (45)
Vyo= Vo Sin (45)

Time for fish to fall the 3.44 m = time for fish to jump 3.44 m

Yv=0=3.44 - 1/2 (9.8) t^2
-3.44=-4.9*t^2
-3.44/-4.9= t^2
.7020=t^2
t= .8379 s

final velocity = 3.44/ .8379
=4.106 m/s

Vyo=4.106
4.106/sin(45)=Vo
Vo= 5.807

Vxo= 5.807 Cos(45)= 4.106

2. Sep 23, 2008

### mal4mac

The final velocity is *not* d/t. Remember the velocity varies over this distance.

You say "Time for fish to fall the 3.44 m = time for fish to jump 3.44 m". Are you sure? I don't see why you need to bother considering the time for fish to fall. Can you do everything just considering the upward motion? Hint:

Uy - vertical initial velocity as fish emerges from water, then:

Uy = gt

3.44 = Uyt - 1/2gt2