A stone is projected at time `t=0` with a speed `v_(0)` at an angle `theta` with the horizontal in a uniform gravitational field. The rate of work done (P) by the gravitational force plotted against time (t) will be as.

Options

A.
B.
C.
D.

Answer

To solve the problem of determining the rate of work done (P) by the gravitational force plotted against time (t) for a stone projected at an angle θ with an initial speed \( v_0 \), we will follow these steps:

### Step 1: Understand the Motion of the Stone
The stone is projected at an angle θ with respect to the horizontal. The motion can be broken down into horizontal and vertical components. The vertical component of the initial velocity is given by:
\[ v_{0y} = v_0 \sin \theta \]
The horizontal component is:
\[ v_{0x} = v_0 \cos \theta \]

### Step 2: Determine the Height (h) as a Function of Time (t)
The height \( h \) of the stone at any time \( t \) can be calculated using the kinematic equation for vertical motion:
\[ h = v_{0y} t - \frac{1}{2} g t^2 \]
Substituting \( v_{0y} \):
\[ h = v_0 \sin \theta \cdot t - \frac{1}{2} g t^2 \]

### Step 3: Calculate the Work Done by Gravitational Force
The work done \( W \) by the gravitational force when the stone is at height \( h \) is given by:
\[ W = -mg h \]
Substituting the expression for \( h \):
\[ W = -mg \left( v_0 \sin \theta \cdot t - \frac{1}{2} g t^2 \right) \]
\[ W = -mg v_0 \sin \theta \cdot t + \frac{1}{2} mg g t^2 \]

### Step 4: Determine the Power (P)
Power \( P \) is defined as the rate of work done with respect to time:
\[ P = \frac{dW}{dt} \]
Calculating the derivative:
\[ P = -mg v_0 \sin \theta + mg g t \]
This simplifies to:
\[ P = mg g t - mg v_0 \sin \theta \]

### Step 5: Analyze the Expression for Power
The expression for power \( P \) can be rewritten as:
\[ P = mg \left( g t - v_0 \sin \theta \right) \]
This indicates that power is a linear function of time \( t \) with a slope of \( mg \) and a y-intercept of \( -mg v_0 \sin \theta \).

### Step 6: Graph the Power vs. Time
The graph of power \( P \) against time \( t \) will be a straight line:
- It will start below zero (negative y-intercept).
- It will have a positive slope (since \( mg > 0 \)).

### Conclusion
The rate of work done (P) by the gravitational force plotted against time (t) will be a straight line with a negative y-intercept and a positive slope.