In the above question, the average power delivered by gravity is

Options

A. (a) `-mg u cos alpha`
B. (b) `-mgu sin alpha`
C. (c) `-(mgucos alpha)/(2)`
D. (d) `-(mgusinalpha)/(2)`

Explanation

To find the average power delivered by gravity when a particle of mass \( m \) is projected at an angle \( \alpha \) to the horizontal with an initial velocity \( u \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Average Power**:
The average power \( P \) delivered by gravity is defined as the total work done by gravity divided by the total time taken to do that work.
\[ P = \frac{W}{t} \]

2. **Calculate the Work Done by Gravity**:
The work done by gravity \( W \) when the particle reaches its maximum height \( h \) is given by:
\[ W = -mgh \]
where \( h \) is the maximum height reached by the particle.

3. **Determine the Maximum Height**:
The maximum height \( h \) can be calculated using the formula:
\[ h = \frac{u^2 \sin^2 \alpha}{2g} \]

4. **Substitute the Maximum Height into the Work Done**:
Substitute \( h \) into the work done equation:
\[ W = -mg \left(\frac{u^2 \sin^2 \alpha}{2g}\right) = -\frac{m u^2 \sin^2 \alpha}{2} \]

5. **Calculate the Time Taken to Reach Maximum Height**:
The time \( t \) taken to reach the maximum height is given by:
\[ t = \frac{u \sin \alpha}{g} \]

6. **Substitute Work Done and Time into the Average Power Formula**:
Now substitute \( W \) and \( t \) into the average power formula:
\[ P = \frac{-\frac{m u^2 \sin^2 \alpha}{2}}{\frac{u \sin \alpha}{g}} \]

7. **Simplify the Expression**:
Simplifying the above expression:
\[ P = -\frac{m u^2 \sin^2 \alpha}{2} \cdot \frac{g}{u \sin \alpha} = -\frac{m g u \sin \alpha}{2} \]

8. **Final Result**:
The average power delivered by gravity is:
\[ P = -\frac{m g u \sin \alpha}{2} \]