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 solve the problem of finding the average power of a projectile projected with an initial speed \( u \) at an angle \( \theta \) from the horizontal, we can follow these steps:

### Step 1: Understanding Average Power
Average power is defined as the total work done divided by the total time taken. Mathematically, it can be expressed as:
\[ P_{\text{avg}} = \frac{W_{\text{total}}}{t_{\text{total}}} \]

### Step 2: Analyzing the Motion
The projectile's motion can be divided into two components: horizontal (x-axis) and vertical (y-axis).

### Step 3: Work Done in the Horizontal Direction
In the horizontal direction, there is no net force acting on the projectile (ignoring air resistance). Since there is no acceleration, the work done \( W_x \) in the horizontal direction is:
\[ W_x = F_x \cdot d_x = 0 \quad (\text{since } F_x = 0) \]

### Step 4: Work Done in the Vertical Direction
In the vertical direction, the projectile returns to its original height after its flight. The initial vertical displacement is \( h \), but since it returns to the same height, the total vertical displacement \( d_y \) is:
\[ d_y = 0 \]
Thus, the work done \( W_y \) in the vertical direction is:
\[ W_y = F_y \cdot d_y = 0 \quad (\text{since } d_y = 0) \]

### Step 5: Total Work Done
The total work done \( W_{\text{total}} \) is the sum of work done in both directions:
\[ W_{\text{total}} = W_x + W_y = 0 + 0 = 0 \]

### Step 6: Total Time of Flight
The total time of flight \( t_{\text{total}} \) for a projectile can be calculated using the formula:
\[ t_{\text{total}} = \frac{2u \sin \theta}{g} \]
where \( g \) is the acceleration due to gravity.

### Step 7: Calculate Average Power
Now substituting the total work done and total time into the average power formula:
\[ P_{\text{avg}} = \frac{W_{\text{total}}}{t_{\text{total}}} = \frac{0}{\frac{2u \sin \theta}{g}} = 0 \]

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