A particle is projected with a velocity `u` making an angle `theta` with the horizontal. The instantaneous power of the gravitational force

Options

A. (a) Varies linearly with time
B. (b) Is constant throughout
C. (c) Is negative for complete path
D. (d) None of the above

Explanation

To find the instantaneous power of the gravitational force acting on a particle projected with an initial velocity \( u \) at an angle \( \theta \) with the horizontal, we can follow these steps:

### Step 1: Identify the Components of Velocity
The initial velocity \( u \) can be broken down into its horizontal and vertical components:
- Horizontal component: \( V_x = u \cos \theta \)
- Vertical component: \( V_y = u \sin \theta - gt \) (where \( g \) is the acceleration due to gravity and \( t \) is the time elapsed)

### Step 2: Write the Velocity Vector
The velocity vector \( \mathbf{V} \) at any time \( t \) can be expressed as:
\[ \mathbf{V} = V_x \hat{i} + V_y \hat{j} = u \cos \theta \hat{i} + (u \sin \theta - gt) \hat{j} \]

### Step 3: Identify the Force Vector
The gravitational force \( \mathbf{F} \) acting on the particle is:
\[ \mathbf{F} = -mg \hat{j} \]

### Step 4: Calculate Instantaneous Power
The instantaneous power \( P \) is given by the dot product of the force vector and the velocity vector:
\[ P = \mathbf{F} \cdot \mathbf{V} \]
Substituting the expressions for \( \mathbf{F} \) and \( \mathbf{V} \):
\[ P = (-mg \hat{j}) \cdot (u \cos \theta \hat{i} + (u \sin \theta - gt) \hat{j}) \]

### Step 5: Evaluate the Dot Product
The dot product simplifies to:
\[ P = -mg \cdot (u \sin \theta - gt) \]
This can be rearranged to:
\[ P = mg(gt - u \sin \theta) \]

### Step 6: Analyze the Result
From the expression \( P = mg(gt - u \sin \theta) \), we can see that the power varies linearly with time \( t \) since \( gt \) is a linear term and \( u \sin \theta \) is a constant.

### Conclusion
Thus, the instantaneous power of the gravitational force acting on the particle varies linearly with time.

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