A body is projected at an angle `alpha` with velocity `32 ft//"sec"`. Its direction of motion will be `alpha//2` after time ………..

Answer

To solve the problem, we need to find the time \( t \) at which the direction of motion of a projectile, initially launched at an angle \( \alpha \) with a velocity of \( 32 \, \text{ft/s} \), becomes \( \frac{\alpha}{2} \) with respect to the horizontal.

### Step-by-Step Solution:

1. **Initial Velocity Components**:
- The initial velocity \( v_0 = 32 \, \text{ft/s} \).
- The horizontal component of the velocity \( v_{x} = v_0 \cos(\alpha) = 32 \cos(\alpha) \).
- The vertical component of the velocity \( v_{y} = v_0 \sin(\alpha) = 32 \sin(\alpha) \).

2. **Vertical Motion**:
- The vertical motion is influenced by gravity, which acts downwards with an acceleration \( g = 32 \, \text{ft/s}^2 \).
- The vertical velocity at time \( t \) is given by:
\[ v_{y} = v_{0} \sin(\alpha) - g t = 32 \sin(\alpha) - 32 t \]

3. **Condition for Direction of Motion**:
- The direction of motion will be \( \frac{\alpha}{2} \) when the tangent of the angle is equal to the ratio of the vertical and horizontal components of the velocity:
\[ \tan\left(\frac{\alpha}{2}\right) = \frac{v_{y}}{v_{x}} \]
- Substituting the expressions for \( v_{y} \) and \( v_{x} \):
\[ \tan\left(\frac{\alpha}{2}\right) = \frac{32 \sin(\alpha) - 32 t}{32 \cos(\alpha)} \]
- Simplifying this gives:
\[ \tan\left(\frac{\alpha}{2}\right) = \frac{\sin(\alpha) - t}{\cos(\alpha)} \]

4. **Using Trigonometric Identity**:
- We know that:
\[ \tan\left(\frac{\alpha}{2}\right) = \frac{\sin(\alpha)}{1 + \cos(\alpha)} \]
- Therefore, we can equate:
\[ \frac{\sin(\alpha)}{1 + \cos(\alpha)} = \frac{\sin(\alpha) - t}{\cos(\alpha)} \]

5. **Cross-Multiplying**:
- Cross-multiplying gives:
\[ \sin(\alpha) \cos(\alpha) = (1 + \cos(\alpha))(\sin(\alpha) - t) \]
- Expanding the right side:
\[ \sin(\alpha) \cos(\alpha) = \sin(\alpha) + \sin(\alpha) \cos(\alpha) - t - t \cos(\alpha) \]
- Simplifying leads to:
\[ t + t \cos(\alpha) = \sin(\alpha) \]

6. **Solving for \( t \)**:
- Factoring out \( t \):
\[ t(1 + \cos(\alpha)) = \sin(\alpha) \]
- Thus, we have:
\[ t = \frac{\sin(\alpha)}{1 + \cos(\alpha)} \]

### Final Answer:
The time \( t \) at which the direction of motion becomes \( \frac{\alpha}{2} \) is:
\[ t = \frac{\sin(\alpha)}{1 + \cos(\alpha)} \]