A particle is projected at angle `theta` with horizontal from ground. The slop (m) of the trajectory of the particle varies with time (t) as

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Answer

To solve the problem of how the slope \( m \) of the trajectory of a particle projected at an angle \( \theta \) with the horizontal varies with time \( t \), we can follow these steps:

### Step 1: Understand the motion components
A particle is projected at an initial speed \( u \) at an angle \( \theta \). The horizontal and vertical components of the initial velocity are:
- Horizontal component: \( u_x = u \cos \theta \)
- Vertical component: \( u_y = u \sin \theta \)

### Step 2: Write the equations of motion
The equations of motion for the particle can be described as follows:
- Horizontal motion:
\[ x = u \cos \theta \cdot t \]
- Vertical motion:
\[ y = u \sin \theta \cdot t - \frac{1}{2} g t^2 \]

### Step 3: Derive the equation of the trajectory
To find the trajectory equation, we eliminate time \( t \) from the equations. From the horizontal motion equation, we can express \( t \) as:
\[ t = \frac{x}{u \cos \theta} \]
Substituting this into the vertical motion equation gives:
\[ y = u \sin \theta \cdot \left(\frac{x}{u \cos \theta}\right) - \frac{1}{2} g \left(\frac{x}{u \cos \theta}\right)^2 \]
Simplifying this, we get:
\[ y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta} \]

### Step 4: Find the slope \( m \)
The slope \( m \) of the trajectory at any point is given by:
\[ m = \frac{dy}{dx} \]
Differentiating the trajectory equation:
\[ m = \tan \theta - \frac{g}{u^2 \cos^2 \theta} x \]

### Step 5: Express \( x \) in terms of \( t \)
From the horizontal motion equation, we know:
\[ x = u \cos \theta \cdot t \]
Substituting this into the slope equation gives:
\[ m = \tan \theta - \frac{g}{u^2 \cos^2 \theta} (u \cos \theta \cdot t) \]
This simplifies to:
\[ m = \tan \theta - \frac{g}{u \cos \theta} t \]

### Step 6: Analyze the slope equation
The equation \( m = \tan \theta - \frac{g}{u \cos \theta} t \) represents a straight line with:
- A y-intercept of \( \tan \theta \)
- A negative slope of \( -\frac{g}{u \cos \theta} \)

### Step 7: Conclusion
The slope \( m \) decreases linearly with time \( t \), indicating that as time progresses, the slope of the trajectory becomes less steep.

### Final Answer
The graph representing the slope of the trajectory of the particle as a function of time \( t \) is a straight line with a negative slope and a positive intercept.

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