A bird is flying with a speed of 12.0 m/s over water when it accidentally drops a 2.30 kg fish.
The acceleration of gravity is 9.81 m/s2 .
If the altitude of the bird is 8.20 m and air resistance is disregarded, what is the speed of the fish when it hits the water?
Answer in units of m/s.
The speed of the fish at water is a vector combination of the intial horizontal speed, and the terminal vertical speed. I will be happy to check your work.
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To find the speed of the fish when it hits the water, we can use the principle of conservation of energy.
The potential energy of the fish when it is at an altitude of 8.20 m is given by the formula:
Potential Energy = mass * acceleration due to gravity * altitude
Plugging in the values, we get:
Potential Energy = 2.30 kg * 9.81 m/s^2 * 8.20 m
Next, we need to equate this potential energy to the kinetic energy of the fish just before it hits the water. The kinetic energy of an object is given by the formula:
Kinetic Energy = (1/2) * mass * velocity^2
We can assume that the velocity of the fish just before it hits the water is v m/s. So the kinetic energy of the fish is given by:
Kinetic Energy = (1/2) * 2.30 kg * v^2
Since energy is conserved, we can equate the potential energy to the kinetic energy:
Potential Energy = Kinetic Energy
2.30 kg * 9.81 m/s^2 * 8.20 m = (1/2) * 2.30 kg * v^2
Now we can solve for v. First, simplify the equation by canceling out the common factors:
9.81 m/s^2 * 8.20 m = (1/2) * v^2
Multiply the numbers on the left side of the equation:
80.262 m^2/s^2 = (1/2) * v^2
Multiply both sides of the equation by 2 to get rid of the fraction:
160.524 m^2/s^2 = v^2
To isolate v, take the square root of both sides:
v = √(160.524) m/s
Calculating the square root, we get:
v ≈ 12.673 m/s
Therefore, the speed of the fish when it hits the water is approximately 12.673 m/s.