A particle is projected with speed u at angle `theta` with horizontal from ground . If it is at same height from ground at time `t_(1) and t_(2)` , then its average velocity in time interval `t_(1)` to `t_(2)` is

Options

A. Zero
B. u sin `theta`
C. u cos `theta`
D. `(1)/(2) [ " u cos " theta] `

Answer

To solve the problem, we need to find the average velocity of a particle projected at an angle θ with an initial speed u, when it is at the same height at two different times, t₁ and t₂.

### Step-by-Step Solution:

1. **Understanding the Motion**:
The particle is projected with an initial speed u at an angle θ. The motion can be analyzed in two components: horizontal (x-direction) and vertical (y-direction).

2. **Components of Initial Velocity**:
- The horizontal component of the initial velocity (u_x) is given by:
\[ u_x = u \cos \theta \]
- The vertical component of the initial velocity (u_y) is given by:
\[ u_y = u \sin \theta \]

3. **Vertical Motion**:
The vertical motion is influenced by gravity (g). The vertical position (y) of the particle at any time t can be described by:
\[ y(t) = u_y t - \frac{1}{2} g t^2 \]

4. **Same Height Condition**:
The particle is at the same height at times t₁ and t₂. This means:
\[ y(t_1) = y(t_2) \]
Since the vertical motion is symmetrical, the time taken to reach the same height on the way up (t₁) and on the way down (t₂) is such that:
\[ t_2 = T - t_1 \]
where T is the total time of flight.

5. **Average Velocity Calculation**:
The average velocity (v_avg) over the time interval from t₁ to t₂ is defined as:
\[ v_{avg} = \frac{\Delta x}{\Delta t} \]
where Δx is the change in horizontal position and Δt is the time interval (t₂ - t₁).

6. **Horizontal Motion**:
The horizontal displacement (x) at any time t is given by:
\[ x(t) = u_x t = (u \cos \theta) t \]
Thus, the horizontal positions at t₁ and t₂ are:
\[ x(t_1) = u \cos \theta \cdot t_1 \]
\[ x(t_2) = u \cos \theta \cdot t_2 \]

7. **Change in Horizontal Position**:
The change in horizontal position (Δx) is:
\[ \Delta x = x(t_2) - x(t_1) = u \cos \theta (t_2 - t_1) \]

8. **Time Interval**:
The time interval (Δt) is:
\[ \Delta t = t_2 - t_1 \]

9. **Substituting into Average Velocity**:
Now substituting Δx and Δt into the average velocity formula:
\[ v_{avg} = \frac{u \cos \theta (t_2 - t_1)}{t_2 - t_1} = u \cos \theta \]

### Final Result:
Thus, the average velocity of the particle in the time interval from t₁ to t₂ is:
\[ \boxed{u \cos \theta} \]