A 5 kg body is fired vertically up with a speed of 200 m/s. Just before it hits the ground, its speed is 150 m/s. Over the entire trip,

Options

A. Change in internal energy of the projectile and air is `-43750 J.`
B. The work done by gravity is 44000 J
C. The work done by gravity is zero
D. The time of upward motion is less than that for downward motion

Answer

To solve the problem, we will analyze the motion of the body fired vertically upward and then coming back down, taking into account the effects of gravity and air resistance.

### Step-by-Step Solution:

1. **Identify the Initial and Final Conditions:**
- The mass of the body, \( m = 5 \, \text{kg} \).
- Initial speed when fired upwards, \( u = 200 \, \text{m/s} \).
- Final speed just before hitting the ground, \( v = 150 \, \text{m/s} \).

2. **Calculate the Change in Kinetic Energy:**
- The initial kinetic energy (KE_initial) when the body is fired upwards can be calculated using the formula:
\[ KE_{\text{initial}} = \frac{1}{2} m u^2 = \frac{1}{2} \times 5 \times (200)^2 = 1000000 \, \text{J} \]
- The final kinetic energy (KE_final) just before hitting the ground is:
\[ KE_{\text{final}} = \frac{1}{2} m v^2 = \frac{1}{2} \times 5 \times (150)^2 = 562500 \, \text{J} \]

3. **Calculate the Change in Kinetic Energy:**
- The change in kinetic energy (ΔKE) is given by:
\[ \Delta KE = KE_{\text{final}} - KE_{\text{initial}} = 562500 - 1000000 = -437500 \, \text{J} \]
- This negative value indicates that the body has lost kinetic energy during its trip.

4. **Work Done by Air Resistance:**
- The work done by air resistance (W_air) can be equated to the change in kinetic energy:
\[ W_{\text{air}} = \Delta KE = -437500 \, \text{J} \]
- This means that air resistance has done negative work on the body, indicating energy loss.

5. **Work Done by Gravity:**
- Since the body returns to the same height from which it was fired, the total displacement is zero. Therefore, the work done by gravity (W_gravity) is:
\[ W_{\text{gravity}} = 0 \, \text{J} \]

6. **Conclusion about Internal Energy:**
- The change in internal energy of the system (projectile and air) is not affected by the work done by gravity since it is zero. The loss in kinetic energy is due to air resistance, which means the internal energy change is also zero.

### Final Answer:
- The work done by gravity is zero.