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.

Options

A. `theta=30^@`
B. `theta=45^@`
C. `theta=60^@`
D. `theta=90^@`

Explanation

To solve the problem, we need to determine the angle of projection of a projectile given the ratio of its velocities at two different heights. Here’s a step-by-step solution:

### Step 1: Understand the Problem
We are given that the velocity of the projectile at its greatest height (v1) is \(\sqrt{\frac{2}{5}}\) times its velocity at half of its greatest height (v2). We need to find the angle of projection \(\theta\).

### Step 2: Define the Velocities
1. At the greatest height, the vertical component of the velocity (v1y) is 0. Therefore, the velocity at this point is:
\[ v_1 = u \cos \theta \]
where \(u\) is the initial velocity and \(\theta\) is the angle of projection.

2. At half of the greatest height, we need to find the vertical component of the velocity (v2y). The maximum height \(H\) of the projectile is given by:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
Thus, half of the maximum height is:
\[ \frac{H}{2} = \frac{u^2 \sin^2 \theta}{4g} \]

### Step 3: Calculate the Vertical Velocity at Half Height
Using the kinematic equation:
\[ v_{y}^2 = u_{y}^2 + 2a s \]
where \(u_{y} = u \sin \theta\), \(a = -g\), and \(s = \frac{H}{2}\):
\[ v_{2y}^2 = (u \sin \theta)^2 - 2g \cdot \frac{u^2 \sin^2 \theta}{4g} \]
This simplifies to:
\[ v_{2y}^2 = u^2 \sin^2 \theta - \frac{u^2 \sin^2 \theta}{2} = \frac{u^2 \sin^2 \theta}{2} \]
Thus:
\[ v_{2y} = \frac{u \sin \theta}{\sqrt{2}} \]

### Step 4: Calculate the Total Velocity at Half Height
The horizontal component of the velocity remains the same:
\[ v_{2x} = u \cos \theta \]
The total velocity at half height (v2) is:
\[ v_2 = \sqrt{(v_{2x})^2 + (v_{2y})^2} = \sqrt{(u \cos \theta)^2 + \left(\frac{u \sin \theta}{\sqrt{2}}\right)^2} \]
This simplifies to:
\[ v_2 = \sqrt{u^2 \cos^2 \theta + \frac{u^2 \sin^2 \theta}{2}} = u \sqrt{\cos^2 \theta + \frac{\sin^2 \theta}{2}} \]

### Step 5: Set Up the Ratio of Velocities
According to the problem:
\[ \frac{v_1}{v_2} = \sqrt{\frac{2}{5}} \]
Substituting the expressions for \(v_1\) and \(v_2\):
\[ \frac{u \cos \theta}{u \sqrt{\cos^2 \theta + \frac{\sin^2 \theta}{2}}} = \sqrt{\frac{2}{5}} \]
This simplifies to:
\[ \frac{\cos \theta}{\sqrt{\cos^2 \theta + \frac{\sin^2 \theta}{2}}} = \sqrt{\frac{2}{5}} \]

### Step 6: Square Both Sides
Squaring both sides gives:
\[ \frac{\cos^2 \theta}{\cos^2 \theta + \frac{\sin^2 \theta}{2}} = \frac{2}{5} \]
Cross-multiplying yields:
\[ 5 \cos^2 \theta = 2 \left(\cos^2 \theta + \frac{\sin^2 \theta}{2}\right) \]
This simplifies to:
\[ 5 \cos^2 \theta = 2 \cos^2 \theta + \sin^2 \theta \]
Using \(\sin^2 \theta = 1 - \cos^2 \theta\):
\[ 5 \cos^2 \theta = 2 \cos^2 \theta + 1 - \cos^2 \theta \]
\[ 5 \cos^2 \theta = \cos^2 \theta + 1 \]
\[ 4 \cos^2 \theta = 1 \]
\[ \cos^2 \theta = \frac{1}{4} \]
Thus:
\[ \cos \theta = \frac{1}{2} \quad \text{(since } \theta \text{ is in the first quadrant)} \]
Therefore:
\[ \theta = 60^\circ \]

### Final Answer
The angle of projection \(\theta\) is \(60^\circ\).

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