If a projectile is projected with initial speed u and angle `theta` from horizontal then what will be its average power up to time when it will hit the ground again ?

Options

A. 100
B. 200
C. 300
D. 0

Answer

To find the average power of a projectile projected with an initial speed \( u \) at an angle \( \theta \) when it hits the ground again, we can follow these steps:

### Step 1: Understand the Concept of Power
Power is defined as the rate of doing work. Mathematically, it can be expressed as:
\[ P = \frac{W}{T} \]
where \( W \) is the work done and \( T \) is the time taken.

### Step 2: Analyze the Forces Acting on the Projectile
In projectile motion, the only force acting on the projectile (after it is launched) is the gravitational force \( mg \) acting downwards. In the horizontal direction, there is no force acting on the projectile (assuming no air resistance).

### Step 3: Determine the Displacement
When the projectile is launched and returns to the ground, the vertical displacement \( y \) is zero because it starts and ends at the same height. Therefore, the total work done in the vertical direction is:
\[ W_y = F_y \cdot d_y = mg \cdot 0 = 0 \]
where \( d_y \) is the vertical displacement.

### Step 4: Calculate the Average Power in the Vertical Direction
Since the vertical displacement is zero, the average velocity in the vertical direction is also zero:
\[ \text{Average Velocity in y} = \frac{\text{Total Displacement in y}}{\text{Total Time}} = \frac{0}{T} = 0 \]
Thus, the average power in the vertical direction is:
\[ P_{avg, y} = F_y \cdot \text{Average Velocity in y} = mg \cdot 0 = 0 \]

### Step 5: Analyze the Horizontal Motion
In the horizontal direction, the projectile moves with a constant velocity \( u \cos \theta \) since there is no horizontal force acting on it. However, since there is no work done in the horizontal direction (as the displacement is not changing the height), the average power in the horizontal direction is also zero:
\[ P_{avg, x} = 0 \]

### Step 6: Combine the Results
Since both the average power in the vertical direction and the horizontal direction are zero, the total average power is:
\[ P_{avg} = P_{avg, x} + P_{avg, y} = 0 + 0 = 0 \]

### Final Answer
The average power of the projectile up to the time it hits the ground again is:
\[ \boxed{0} \]