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

Options

A. `30^@`
B. `60^@`
C. `45^@`
D. `0^@`

Answer

To solve the problem, we need to analyze the speeds of the projectile at its maximum height and at half the maximum height. Let's denote the initial speed of the projectile as \( u \) and the angle of projection as \( \theta \).

### Step 1: Determine the speed at maximum height
At the maximum height, the vertical component of the velocity becomes zero, and only the horizontal component remains. The horizontal component of the initial velocity is given by:
\[ V_{x} = u \cos \theta \]
Thus, the speed at maximum height \( V_h \) is:
\[ V_h = u \cos \theta \]

### Step 2: Determine the height of the projectile
The maximum height \( H \) of the projectile can be calculated using the formula:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
where \( g \) is the acceleration due to gravity.

### Step 3: Determine the speed at half the maximum height
Half of the maximum height is:
\[ h = \frac{H}{2} = \frac{u^2 \sin^2 \theta}{4g} \]
At this height, the vertical component of the velocity can be calculated using the kinematic equation:
\[ V_{y} = \sqrt{u^2 \sin^2 \theta - 2gh} \]
Substituting for \( h \):
\[ V_{y} = \sqrt{u^2 \sin^2 \theta - 2g \left(\frac{u^2 \sin^2 \theta}{4g}\right)} = \sqrt{u^2 \sin^2 \theta - \frac{u^2 \sin^2 \theta}{2}} = \sqrt{\frac{u^2 \sin^2 \theta}{2}} = \frac{u \sin \theta}{\sqrt{2}} \]
The horizontal component remains the same:
\[ V_{x} = u \cos \theta \]
Thus, the speed at half the maximum height \( V_{h/2} \) is:
\[ V_{h/2} = \sqrt{V_{x}^2 + V_{y}^2} = \sqrt{(u \cos \theta)^2 + \left(\frac{u \sin \theta}{\sqrt{2}}\right)^2} \]
\[ = \sqrt{u^2 \cos^2 \theta + \frac{u^2 \sin^2 \theta}{2}} = u \sqrt{\cos^2 \theta + \frac{\sin^2 \theta}{2}} \]

### Step 4: Set up the equation based on the given condition
According to the problem, the speed at maximum height is \( \frac{\sqrt{2}}{5} \) times the speed at half the maximum height:
\[ u \cos \theta = \frac{\sqrt{2}}{5} V_{h/2} \]
Substituting \( V_{h/2} \):
\[ u \cos \theta = \frac{\sqrt{2}}{5} \left( u \sqrt{\cos^2 \theta + \frac{\sin^2 \theta}{2}} \right) \]
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 5: Square both sides to eliminate the square root
\[ \cos^2 \theta = \frac{2}{25} \left( \cos^2 \theta + \frac{\sin^2 \theta}{2} \right) \]
Multiplying through by 25:
\[ 25 \cos^2 \theta = 2 \left( \cos^2 \theta + \frac{\sin^2 \theta}{2} \right) \]
This simplifies to:
\[ 25 \cos^2 \theta = 2 \cos^2 \theta + \sin^2 \theta \]
Using \( \sin^2 \theta = 1 - \cos^2 \theta \):
\[ 25 \cos^2 \theta = 2 \cos^2 \theta + 1 - \cos^2 \theta \]
\[ 25 \cos^2 \theta = (2 - 1) \cos^2 \theta + 1 \]
\[ 25 \cos^2 \theta = \cos^2 \theta + 1 \]
Rearranging gives:
\[ 24 \cos^2 \theta = 1 \]
\[ \cos^2 \theta = \frac{1}{24} \]
Taking the square root:
\[ \cos \theta = \frac{1}{\sqrt{24}} = \frac{1}{2\sqrt{6}} \]

### Step 6: Find \( \theta \)
Using the relationship \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ \sin^2 \theta = 1 - \frac{1}{24} = \frac{23}{24} \]
Thus, \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). Using the values:
\[ \tan^2 \theta = \frac{\frac{23}{24}}{\frac{1}{24}} = 23 \]
Therefore, \( \theta = \tan^{-1}(\sqrt{23}) \).

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