Finding speed when fish jumps out of water

[sydneyown]

Homework Statement

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.

Homework Equations

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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

it told me that it was the wrong answer. please help!

 

[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

 

[baba89]

Your approach to solving this problem seems correct. However, there may be a mistake in your calculation for the final velocity. The correct equation to use is Vf = Vo + at, where Vf is the final velocity, Vo is the initial velocity, a is the acceleration (in this case, gravity), and t is the time. Using this equation, we get:

Vf = 0 + (-9.8)(0.8379) = -8.204 m/s

Since the fish is jumping upwards, the final velocity should be positive, not negative. This means that your initial velocity calculation for the y-direction (Vyo) is also incorrect. The correct calculation would be:

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

Therefore, the fish's speed on emerging from the water would be:

V = √(Vx^2 + Vy^2) = √(4.106^2 + 4.106^2) = 5.807 m/s

This matches your initial velocity calculation for the x-direction, which is correct.

In summary, the mistake in your initial calculation for the final velocity (Vf) resulted in incorrect values for both Vyo and the final speed. By using the correct equation, you should be able to arrive at the correct answer.