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 θ with the horizontal varies with time (t), we can follow these steps:

### Step 1: Understand the Motion Components
The particle is projected with an initial speed \( u \) at an angle \( \theta \). We can break this motion into horizontal and vertical components:
- Horizontal component of velocity: \( u_x = u \cos \theta \)
- Vertical component of velocity: \( u_y = u \sin \theta \)

### Step 2: Write the Equations of Motion
For horizontal motion (x-direction):
- There is no acceleration, so the position as a function of time is given by:
\[ x = u_x t = u \cos \theta \cdot t \]

For vertical motion (y-direction):
- The vertical position as a function of time is given by:
\[ y = u_y t - \frac{1}{2} g t^2 = u \sin \theta \cdot t - \frac{1}{2} g t^2 \]

### Step 3: Eliminate Time to Find the Trajectory Equation
We can express time \( t \) in terms of \( x \) from the horizontal motion equation:
\[ t = \frac{x}{u \cos \theta} \]

Substituting this expression for \( t \) into the vertical motion equation:
\[ y = u \sin \theta \left(\frac{x}{u \cos \theta}\right) - \frac{1}{2} g \left(\frac{x}{u \cos \theta}\right)^2 \]
This simplifies to:
\[ y = x \tan \theta - \frac{g}{2 u^2 \cos^2 \theta} x^2 \]
This is the equation of the trajectory.

### Step 4: Determine the Slope of the Trajectory
The slope \( m \) of the trajectory at any point is given by the derivative \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \tan \theta - \frac{g}{u^2 \cos^2 \theta} x \]
This shows that the slope \( m \) is a linear function of \( x \).

### Step 5: Relate Slope to Time
To express \( m \) as a function of time, we substitute \( x = u \cos \theta \cdot t \) into the slope equation:
\[ m = \tan \theta - \frac{g}{u^2 \cos^2 \theta} (u \cos \theta \cdot t) \]
This simplifies to:
\[ m = \tan \theta - \frac{g}{u \cos \theta} t \]
Thus, the slope \( m \) varies linearly with time \( t \).

### Final Result
The slope \( m \) of the trajectory as a function of time \( t \) is given by:
\[ m(t) = \tan \theta - \frac{g}{u \cos \theta} t \]