# Sea Bird & Fish Question

1. Sep 13, 2014

### IronManTable

1. The problem statement, all variables and given/known data

A sea bird catches a fish and then flies in an easterly direction with a speed v=7.39 m/s, h = 4.69 m above the surface of the sea. The fish has a mass of 115.3 g and is still alive. It thrashes and wiggles and as a result the bird drops the fish.

i) What is the speed of the fish when the bird drops it?

ii) What is the speed of the fish when it hits the water?

iii) When the fish hits the water it is momentarily stunned. It comes to a rest in 0.111 s. What is the magnitude of the force experienced by the fish as it comes to rest?

iv) How far horizontally from where it was dropped does the fish hit the water?

2. Relevant equations

v2 = u2 +2gh
s = ut + 0.5 x t2
F = ma

3. The attempt at a solution

i) What is the speed of the fish when the bird drops it?

Since v1 = 0
v2 = √(2 x 9.8 x 4.69)
v2 = 9.59 m/s

ii) What is the speed of the fish when it hits the water?
Using Pythagoras' Theorem,
a2 + b2 = c2
∴ c = √(7.392 + 9.592)
c = 12.1 m/s

iii) When the fish hits the water it is momentarily stunned. It comes to a rest in 0.111 s. What is the magnitude of the force experienced by the fish as it comes to rest?
s = ut + 0.5 x t2
4.69 = 0 + 0.5 x a x 0.1112
a = 4.69 / (0.5 x 0.1112)
a = 761.3 m/s2
F = ma
F = 0.1153 x 761.3
F = 87.8 N

iv) How far horizontally from where it was dropped does the fish hit the water?

I can't seem to figure out a way to get this answer. Can someone please check if my answers are right and can give me an insight on how to answer this last question? Thank you!​

2. Sep 13, 2014

### Staff: Mentor

[noparse]

(i) the speed of the fish at the moment the bird opens its talons;
(ii) you have applied the correct method, though answer can't be right.
(iii) wrong. Its movement in water is what takes 0.111s.[/noparse]

Last edited: Sep 13, 2014
3. Sep 13, 2014

### IronManTable

So how can I do the ones I got incorrect?

4. Sep 13, 2014

### Staff: Mentor

Try (i) again.

5. Sep 13, 2014

### IronManTable

Okay well I tried it in a new foruma: v2 = v02 + 2ay(y - y0)

v2 = v02 + 2ay(y - y0)
v2 = 0 + 2(-9.8)(0 - 4.69)
v2 = 91.924
v = 9.59 m/s

I come up with the same answer?

6. Sep 13, 2014

### ehild

The fish in the beak of the bird travels horizontally with the speed of the bird, and keeps that horizontal velocity component during its downward fall. The velocity has both horizontal and vertical components....

ehild

7. Sep 13, 2014

### Staff: Mentor

You have just calculated the fish's vertical speed after it has fallen to the water. But that is not what part (i) is asking.

This part of the question asks what is the speed of the fish before it starts to fall.

8. Sep 13, 2014

### IronManTable

Oh! I get it now. It would be the same speed as the bird, correct? So, 7.39 m/s ?

9. Sep 13, 2014

### Staff: Mentor

Right.

Now you have the right data values to do (ii).

10. Sep 13, 2014

### IronManTable

Great! I've done (ii) and (iii) and managed to get the correct values. However, i can't seem to get the final one.

11. Sep 13, 2014

### ehild

The motion of the fish is the resultant of a horizontal motion and a vertical one - it moves like a projectile.

ehild

12. Sep 14, 2014

### CWatters

Hint: With projectile problems there is usually one parameter that is the same for both the horizontal and vertical motion and that is time. The horizontal and vertical velocities may well be different but the object usually starts and stops it's motion in both planes at the same time.