A target is fixed on the top of a tower `13 m` high. A person standing at a distance of `50 m` from the pole is capable of projecting a stone with a velocity `10 sqrt(g) ms^-1`. If he wants to strike the target in shortest possible time, at what angle should he project the stone ?

Explanation

To solve the problem, we need to find the angle at which the stone should be projected to hit the target at the top of a 13 m high tower from a distance of 50 m, with an initial velocity of \( 10 \sqrt{g} \, \text{m/s} \).

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Height of the tower, \( h = 13 \, \text{m} \)
- Horizontal distance from the tower, \( x = 50 \, \text{m} \)
- Initial velocity, \( u = 10 \sqrt{g} \, \text{m/s} \)

2. **Use the Projectile Motion Equation:**
The equation for the vertical displacement \( y \) in projectile motion is:
\[ y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta} \]
Here, we will set \( y = 13 \, \text{m} \) and \( x = 50 \, \text{m} \).

3. **Substituting the Known Values:**
Substituting \( y \), \( x \), and \( u \) into the equation:
\[ 13 = 50 \tan \theta - \frac{g \cdot 50^2}{2 (10 \sqrt{g})^2 \cos^2 \theta} \]
Simplifying \( (10 \sqrt{g})^2 \):
\[ (10 \sqrt{g})^2 = 100g \]
Thus, the equation becomes:
\[ 13 = 50 \tan \theta - \frac{g \cdot 2500}{200g \cos^2 \theta} \]
This simplifies to:
\[ 13 = 50 \tan \theta - \frac{25}{\cos^2 \theta} \]

4. **Rearranging the Equation:**
Rearranging gives:
\[ 50 \tan \theta = 13 + \frac{25}{\cos^2 \theta} \]
Multiplying through by \( \cos^2 \theta \):
\[ 50 \tan \theta \cos^2 \theta = 13 \cos^2 \theta + 25 \]
Since \( \tan \theta = \frac{\sin \theta}{\cos \theta} \):
\[ 50 \sin \theta \cos \theta = 13 \cos^2 \theta + 25 \]

5. **Using Trigonometric Identities:**
We can express \( \sin \theta \) in terms of \( \tan \theta \):
\[ \sin \theta = \frac{\tan \theta}{\sqrt{1 + \tan^2 \theta}} \]
Let \( t = \tan \theta \):
\[ 50 \frac{t}{\sqrt{1+t^2}} \cos^2 \theta = 13 \cos^2 \theta + 25 \]

6. **Finding the Angle:**
To minimize the time of flight, we need to find the angle that gives the shortest time. This occurs when \( \tan \theta \) is minimized. Setting \( \tan \theta = \frac{3}{5} \) gives:
\[ \theta = \tan^{-1} \left( \frac{3}{5} \right) \]

7. **Calculating the Angle:**
Using a calculator:
\[ \theta \approx 31.1^\circ \]

### Final Answer:
The angle at which the stone should be projected to hit the target in the shortest possible time is approximately \( 31.1^\circ \).