1) Seagulls are often observed dropping clams and other shellfish from a height to the rocks below, as a mean of opening the shells. If a seagull drops a shell from rest at a height of 14 m, how fast is the shell moving when it hits the rocks?

1/2 m v ^2 = m g h

To find the speed of the shell when it hits the rocks, we can use the principle of conservation of energy.

Step 1: Determine the potential energy of the shell at the initial height.
The potential energy (PE) of the shell at a height of 14 m can be calculated using the formula:
PE = mgh
where m is the mass of the shell, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height.

Step 2: Determine the kinetic energy of the shell at the point of impact.
The kinetic energy (KE) of the shell at the point of impact will be equal to the potential energy calculated in step 1, as energy is conserved.

Step 3: Equate the potential energy to the kinetic energy to find the speed of the shell.
PE = KE
mgh = (1/2)mv^2

Step 4: Solve the equation for v (the speed of the shell).
v = sqrt(2gh)

Step 5: Substitute the given values into the equation and calculate the speed.
v = sqrt(2 * 9.8 * 14)
v ≈ 17.76 m/s

Therefore, the shell will be moving at approximately 17.76 m/s when it hits the rocks.

To determine the speed of the shell when it hits the rocks, we can use the principle of conservation of energy.

The potential energy of the shell at the beginning (when it is at a height of 14 m) is equal to the kinetic energy of the shell when it hits the rocks. The potential energy (PE) is given by the formula PE = mgh, where m is the mass of the shell, g is the acceleration due to gravity (9.8 m/s^2), and h is the height (14 m).

Now, to find the kinetic energy (KE) of the shell, we can use the formula KE = 1/2mv^2, where v is the speed of the shell when it hits the rocks.

Since the potential energy at the beginning is equal to the kinetic energy when it hits the rocks, we have:

mgh = 1/2mv^2

The mass of the shell cancels out:

gh = 1/2v^2

Now we can solve for v:

v^2 = 2gh

Taking the square root of both sides:

v = √(2gh)

Plugging in the values, with g = 9.8 m/s^2 and h = 14 m:

v = √(2 * 9.8 * 14) ≈ 17.68 m/s

Therefore, the shell will be moving at approximately 17.68 m/s when it hits the rocks. Remember, this is assuming no air resistance.