A particle is projected with a speed v and an angle `theta` to the horizontal. After a time t, the magnitude of the instantaneous velocity is equal to the magnitude of the average velocity from 0 to t. Find t.

Explanation

To solve the problem, we need to find the time \( t \) at which the magnitude of the instantaneous velocity of a projectile is equal to the magnitude of the average velocity from the time of projection to time \( t \).

### Step-by-Step Solution:

1. **Understanding Instantaneous Velocity:**
The instantaneous velocity \( \vec{v}(t) \) of a particle projected with speed \( v \) at an angle \( \theta \) can be expressed as:
\[ \vec{v}(t) = v \cos \theta \hat{i} + (v \sin \theta - gt) \hat{j} \]
The magnitude of this instantaneous velocity is:
\[ |\vec{v}(t)| = \sqrt{(v \cos \theta)^2 + (v \sin \theta - gt)^2} \]

2. **Calculating the Magnitude of Instantaneous Velocity:**
Expanding the expression:
\[ |\vec{v}(t)| = \sqrt{v^2 \cos^2 \theta + (v \sin \theta - gt)^2} \]
\[ = \sqrt{v^2 \cos^2 \theta + (v^2 \sin^2 \theta - 2v \sin \theta gt + g^2 t^2)} \]
\[ = \sqrt{v^2 (\cos^2 \theta + \sin^2 \theta) + g^2 t^2 - 2g t v \sin \theta} \]
Since \( \cos^2 \theta + \sin^2 \theta = 1 \):
\[ = \sqrt{v^2 + g^2 t^2 - 2g t v \sin \theta} \quad \text{(Equation 1)} \]

3. **Calculating Average Velocity:**
The average velocity \( \vec{V}_{avg} \) is defined as the total displacement divided by the total time:
\[ \vec{V}_{avg} = \frac{\Delta \vec{r}}{t} \]
The displacement in the horizontal direction \( x \) is given by:
\[ x = v \cos \theta \cdot t \]
The displacement in the vertical direction \( y \) is given by:
\[ y = v \sin \theta \cdot t - \frac{1}{2} g t^2 \]
Thus, the total displacement \( \Delta \vec{r} \) is:
\[ \Delta \vec{r} = \sqrt{x^2 + y^2} = \sqrt{(v \cos \theta t)^2 + \left(v \sin \theta t - \frac{1}{2} g t^2\right)^2} \]

4. **Calculating the Magnitude of Average Velocity:**
The average velocity magnitude is:
\[ |\vec{V}_{avg}| = \frac{1}{t} \sqrt{(v \cos \theta t)^2 + \left(v \sin \theta t - \frac{1}{2} g t^2\right)^2} \]
Simplifying this:
\[ = \frac{1}{t} \sqrt{v^2 \cos^2 \theta t^2 + \left(v \sin \theta t - \frac{1}{2} g t^2\right)^2} \]

5. **Equating Instantaneous and Average Velocity:**
We set the magnitudes of instantaneous and average velocities equal:
\[ \sqrt{v^2 + g^2 t^2 - 2g t v \sin \theta} = \frac{1}{t} \sqrt{v^2 \cos^2 \theta t^2 + \left(v \sin \theta t - \frac{1}{2} g t^2\right)^2} \]

6. **Squaring Both Sides:**
Squaring both sides to eliminate the square roots and simplifying leads to:
\[ v^2 + g^2 t^2 - 2g t v \sin \theta = \frac{1}{t^2} \left(v^2 \cos^2 \theta t^2 + \left(v \sin \theta t - \frac{1}{2} g t^2\right)^2\right) \]

7. **Solving for \( t \):**
After simplification, you will find:
\[ t = \frac{4v \sin \theta}{3g} \]

### Final Answer:
The time \( t \) after which the magnitude of the instantaneous velocity is equal to the magnitude of the average velocity is:
\[ t = \frac{4v \sin \theta}{3g} \]