A heavy stone is thrown from a cliff of height h in a given direction. The speed with which it hits the ground

Options

A. must depend on the speed of projection
B. must be larger than the speed of projection
C. must be independent of the speed of projection
D. may be smaller than the speed of projection

Answer

To find the speed with which a heavy stone hits the ground after being thrown from a cliff of height \( h \), we can use the work-energy theorem. Here’s a step-by-step solution:

### Step 1: Identify the initial conditions
- Let \( h \) be the height of the cliff.
- Let \( u \) be the initial speed of the stone when it is thrown.
- The stone is thrown at an angle, but for the purpose of finding the speed when it hits the ground, we can consider the vertical component of the motion.

### Step 2: Apply the work-energy theorem
According to the work-energy theorem:
\[ \text{Work done by gravity} = \text{Change in kinetic energy} \]

The work done by gravity when the stone falls a height \( h \) is:
\[ W = mgh \]
where \( m \) is the mass of the stone and \( g \) is the acceleration due to gravity.

### Step 3: Write the change in kinetic energy
The change in kinetic energy is given by:
\[ \Delta KE = KE_{\text{final}} - KE_{\text{initial}} = \frac{1}{2}mv^2 - \frac{1}{2}mu^2 \]

### Step 4: Set up the equation
Setting the work done equal to the change in kinetic energy:
\[ mgh = \frac{1}{2}mv^2 - \frac{1}{2}mu^2 \]

### Step 5: Simplify the equation
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ gh = \frac{1}{2}v^2 - \frac{1}{2}u^2 \]
Multiplying through by 2 gives:
\[ 2gh = v^2 - u^2 \]

### Step 6: Solve for \( v^2 \)
Rearranging the equation to solve for \( v^2 \):
\[ v^2 = u^2 + 2gh \]

### Step 7: Take the square root to find \( v \)
Taking the square root of both sides gives:
\[ v = \sqrt{u^2 + 2gh} \]

### Conclusion
The speed with which the stone hits the ground depends on the initial speed of projection \( u \) and the height \( h \).