A woman stands on the edge of a cliff and throws a stone vertically downward with an initial speed of 10 m/s. The instant before the stone hits the ground below, it has 450 J of kinetic energy. If she were to throw the stone horizontally outward from the cliff with the same initial speed of 10 m/s, how much kinetic energy would it have just before it hits the ground ?

Options

A. 50J
B. 100J
C. 450J
D. 800J

Answer

To solve the problem, we will analyze the two scenarios: throwing the stone vertically downward and throwing it horizontally. We will use the principle of conservation of energy to find the kinetic energy of the stone just before it hits the ground in both cases.

### Step-by-Step Solution:

1. **Understanding the First Case (Vertical Throw)**:
- The woman throws the stone vertically downward with an initial speed of \( v_i = 10 \, \text{m/s} \).
- The kinetic energy (KE) just before it hits the ground is given as \( KE = 450 \, \text{J} \).
- The potential energy (PE) at the height \( h \) when thrown is not given, but it will be zero just before impact since the height at that point is zero.

2. **Conservation of Energy in the First Case**:
- The total mechanical energy at the point of throwing (initial) is the sum of the initial kinetic energy and the potential energy:
\[ E_{\text{initial}} = KE_{\text{initial}} + PE_{\text{initial}} = \frac{1}{2} m v_i^2 + mgh \]
- The total mechanical energy just before hitting the ground (final) is:
\[ E_{\text{final}} = KE_{\text{final}} + PE_{\text{final}} = KE_{\text{final}} + 0 = KE_{\text{final}} \]
- By conservation of energy:
\[ E_{\text{initial}} = E_{\text{final}} \implies KE_{\text{initial}} + PE_{\text{initial}} = KE_{\text{final}} \]
- Since \( KE_{\text{final}} = 450 \, \text{J} \), we can express the initial energy as:
\[ KE_{\text{initial}} + mgh = 450 \]

3. **Understanding the Second Case (Horizontal Throw)**:
- In the second scenario, the woman throws the stone horizontally with the same initial speed \( v_i = 10 \, \text{m/s} \).
- The initial kinetic energy in this case is the same as in the first case:
\[ KE_{\text{initial}} = \frac{1}{2} m v_i^2 \]
- The stone still falls the same height \( h \), so the potential energy at the height \( h \) is the same as in the first case.

4. **Conservation of Energy in the Second Case**:
- The total mechanical energy just before hitting the ground (final) is:
\[ E_{\text{final}} = KE' + PE' = KE' + 0 = KE' \]
- By conservation of energy:
\[ KE_{\text{initial}} + mgh = KE' \]
- Since the initial energy is the same as in the first case, we can substitute:
\[ KE' = 450 \, \text{J} \]

5. **Conclusion**:
- The kinetic energy of the stone just before it hits the ground when thrown horizontally is also \( KE' = 450 \, \text{J} \).

### Final Answer:
The kinetic energy of the stone just before it hits the ground when thrown horizontally is **450 J**.