# Bird dropping fish question

#### HexyRexySmexy

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?

My attempt...

v^2 = u^2 +2gh
v2 ^2 = v2 ^2 + 2ad
s = ut + 0.5 x t^2
F = ma

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

Since v1 = 0
v2 ^2 = v1 ^2 + 2ad
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,
a^2+ b^2 = c^2
∴ c = √(7.39^2 + 9.59^2)
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 t^2
4.69 = 0 + 0.5 x a x 0.111^2
a = 4.69 / (0.5 x 0.111^2)
a = 761.3 m/s^2
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!​

#### HexyRexySmexy

Nevermind about the first 3 questions. I ended up fixing them to the correct answers. However, I can't seem to figure out a way to get the last question. Does anyone have a way? Thanks

#### HexyRexySmexy

Can someone quickly comment whether or not this is correct?

s = ut + 0.5 x t^2
s = (12.1 x 0.111) + 0.5 x (0.111)^2
s = 2.02 metres

Thanks!!

#### ChipB

PHF Helper
For the last question the horizontal distance traveled before the fish hits the water is the horizontal velocity of the fish at release times the time it takes to fall to the water. That time is simply the time required for an object to drop 4.69m.

#### HexyRexySmexy

So, would the answer come from the distance over the speed found in (ii) 12.1 m/s to get the time, then multiplying that time by the initial horizontal speed to get the distance?

4.69 m / 12.1 m/s = 0.39 s
then to find the horizontal distance: 7.39 m/s x 0.39 = 2.86 m ?

Thanks again.

#### ChipB

PHF Helper
That's not right - you are assuming that the rate of fall is constant, but it's not. The time it takes to fall is found from

s= (1/2)gt^2

Where h= 4.69m.