Potential Energy and Kinetic Energy Question

AI Thread Summary
A bird drops a 2.20 kg fish from a height of 7.70 m while flying at 17.4 m/s, and the problem involves calculating the fish's speed upon hitting the water, disregarding air resistance. The approach involves using the potential energy equation (PE = mgh) to find the change in potential energy and converting it to kinetic energy (KE = 1/2 mv^2). It is crucial to account for the fish's initial kinetic energy when it is released, as it already has a speed of 17.4 m/s. By combining the initial kinetic energy of the fish with the potential energy lost during the fall, the correct final speed can be determined. The solution was successfully reached after considering these factors.
gansta344u
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Homework Statement


A bird is flying with a speed of 17.4 m/s over
water when it accidentally drops a 2.20 kg
fish.
The acceleration of gravity is 9.81 m/s2 .
If the altitude of the bird is 7.70 m and air
resistance is disregarded, what is the speed of
the fish when it hits the water? Answer in
units of m/s.


Homework Equations


i tried to do potential energy equation PEg=mgh and then since it is converted to KE i re- arange the equation KE=1/2mv^2 and i tried it and it was wrong

3. The Attempt at a Solution [/b
 
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gansta344u said:

Homework Equations


i tried to do potential energy equation PEg=mgh and then since it is converted to KE i re- arange the equation KE=1/2mv^2 and i tried it and it was wrong
The change in PE will tell you the change in KE. Don't forget that the fish already had KE when it was released.
 
so i have to add the KE of bird and PE of the fish
then use the KE=1/2MV^2
 
gansta344u said:
so i have to add the KE of bird and PE of the fish
then use the KE=1/2MV^2
Add the initial KE of the fish. (How fast is the fish moving when it's released?)
 
thanks a lot,i got it rite this time
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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