The velocity of a projectile when it is at the greatest height is `(sqrt (2//5))` times its velocity when it is at half of its greatest height. Determine its angle of projection.

Explanation

To solve the problem, we need to determine the angle of projection \( \theta \) of a projectile given that the velocity at its greatest height is \( \frac{\sqrt{2}}{5} \) times its velocity at half of its greatest height. Let's break down the solution step by step.

### Step 1: Understand the Motion of the Projectile

When a projectile is launched, it has an initial velocity \( u \) which can be resolved into two components:
- Horizontal component: \( u_x = u \cos \theta \)
- Vertical component: \( u_y = u \sin \theta \)

At the greatest height \( h \), the vertical component of the velocity becomes zero, while the horizontal component remains constant.

### Step 2: Determine the Velocity at Greatest Height \( V_1 \)

At the greatest height, the velocity \( V_1 \) is purely horizontal:
\[ V_1 = u_x = u \cos \theta \]

### Step 3: Determine the Height \( h \)

The maximum height \( h \) reached by the projectile can be calculated using the formula:
\[ h = \frac{u^2 \sin^2 \theta}{2g} \]

### Step 4: Determine the Velocity at Half of the Greatest Height \( V_2 \)

At half of the greatest height \( \frac{h}{2} \), we can use the kinematic equation to find the vertical component of the velocity \( V_{y2} \):
\[ V_{y2}^2 = u_y^2 - 2g \left(\frac{h}{2}\right) \]
Substituting \( h \):
\[ V_{y2}^2 = (u \sin \theta)^2 - 2g \left(\frac{u^2 \sin^2 \theta}{4g}\right) \]
\[ = u^2 \sin^2 \theta - \frac{u^2 \sin^2 \theta}{2} \]
\[ = \frac{u^2 \sin^2 \theta}{2} \]
Thus, the vertical component of the velocity at half the height is:
\[ V_{y2} = \sqrt{\frac{u^2 \sin^2 \theta}{2}} = \frac{u \sin \theta}{\sqrt{2}} \]

The horizontal component remains the same:
\[ V_{x2} = u \cos \theta \]

### Step 5: Determine the Resultant Velocity at Half of the Greatest Height \( V_2 \)

The resultant velocity \( V_2 \) at half the height is given by:
\[ V_2 = \sqrt{V_{x2}^2 + V_{y2}^2} \]
Substituting the components:
\[ V_2 = \sqrt{(u \cos \theta)^2 + \left(\frac{u \sin \theta}{\sqrt{2}}\right)^2} \]
\[ = \sqrt{u^2 \cos^2 \theta + \frac{u^2 \sin^2 \theta}{2}} \]
\[ = u \sqrt{\cos^2 \theta + \frac{\sin^2 \theta}{2}} \]

### Step 6: Set Up the Given Condition

According to the problem, the relationship between \( V_1 \) and \( V_2 \) is:
\[ V_1 = \frac{\sqrt{2}}{5} V_2 \]
Substituting the expressions for \( V_1 \) and \( V_2 \):
\[ u \cos \theta = \frac{\sqrt{2}}{5} \left( u \sqrt{\cos^2 \theta + \frac{\sin^2 \theta}{2}} \right) \]
Dividing both sides by \( u \) (assuming \( u \neq 0 \)):
\[ \cos \theta = \frac{\sqrt{2}}{5} \sqrt{\cos^2 \theta + \frac{\sin^2 \theta}{2}} \]

### Step 7: Square Both Sides

Squaring both sides:
\[ \cos^2 \theta = \frac{2}{25} \left( \cos^2 \theta + \frac{\sin^2 \theta}{2} \right) \]
Multiplying through by 25:
\[ 25 \cos^2 \theta = 2 \left( \cos^2 \theta + \frac{\sin^2 \theta}{2} \right) \]
\[ 25 \cos^2 \theta = 2 \cos^2 \theta + \sin^2 \theta \]
Using \( \sin^2 \theta = 1 - \cos^2 \theta \):
\[ 25 \cos^2 \theta = 2 \cos^2 \theta + 1 - \cos^2 \theta \]
\[ 25 \cos^2 \theta = \cos^2 \theta + 1 \]
\[ 24 \cos^2 \theta = 1 \]
\[ \cos^2 \theta = \frac{1}{24} \]

### Step 8: Find \( \tan \theta \)

Using \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ \sin^2 \theta = 1 - \frac{1}{24} = \frac{23}{24} \]
Now, calculate \( \tan \theta \):
\[ \tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} = \frac{\frac{23}{24}}{\frac{1}{24}} = 23 \]
Thus, \( \tan \theta = \sqrt{23} \).

### Step 9: Find the Angle \( \theta \)

Finally, we can find \( \theta \):
\[ \theta = \tan^{-1}(\sqrt{23}) \]

### Conclusion

The angle of projection \( \theta \) is \( \tan^{-1}(\sqrt{23}) \).