Question

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

Answer 1

To find the speed of the shell, we can use the principles of physics. By equating gravitational potential energy to kinetic energy, we can solve for the speed. The speed of the shell when it hits the rocks is approximately 16.72 m/s.

Step-by-step explanation:

To find the speed of the shell when it hits the rocks, we can use the principles of physics. The initial potential energy of the shell is converted into kinetic energy as it falls. We can use the equation for gravitational potential energy, PE = m * g * h, 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 from which it falls. In this case, the initial potential energy is equal to the final kinetic energy, KE = (1/2) * m * v^2, where v is the speed of the shell when it hits the rocks. By equating the two equations, we can solve for v:

1. First, we find the initial potential energy: PE = m * g * h = m * 9.8 * 14

2. Next, we equate the potential energy to the kinetic energy: PE = KE = (1/2) * m * v^2

3. Simplifying the equation, we get: m * 9.8 * 14 = (1/2) * m * v^2

4. Canceling out the mass on both sides, we have: v^2 = 9.8 * 14 * 2

5. Taking the square root of both sides, we find: v = sqrt(9.8 * 14 * 2)

Calculating the value, we get v ≈ 16.72 m/s.