Find the angle of projection for a projectile if its speed at maximum height is `sqrt(5//9)` times its speed at one-third of maximum height `(g=10ms^(-2))`

Answer

To solve the problem, we need to find the angle of projection \( \theta \) for a projectile given that its speed at maximum height is \( \sqrt{\frac{5}{9}} \) times its speed at one-third of maximum height. We will use the equations of motion and the properties of projectile motion.

### Step 1: Understanding the components of velocity
When a projectile is launched with an initial velocity \( u \) at an angle \( \theta \):
- The horizontal component of velocity is \( u \cos \theta \).
- The vertical component of velocity is \( u \sin \theta \).

At maximum height, the vertical component of velocity becomes zero, so the speed at maximum height is simply the horizontal component:
\[ v_{max\ height} = u \cos \theta \]

### Step 2: Finding the speed at one-third of maximum height
At one-third of the maximum height, the horizontal component remains the same \( u \cos \theta \), but we need to find the vertical component at this height.

Using the equation of motion:
\[ v_y^2 = u_y^2 - 2g h \]
where:
- \( u_y = u \sin \theta \) (initial vertical velocity)
- \( h = \frac{H}{3} \) (where \( H \) is the maximum height)

The maximum height \( H \) is given by:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
Thus, \( h = \frac{H}{3} = \frac{u^2 \sin^2 \theta}{6g} \).

Now substituting this into the equation:
\[ v_y^2 = (u \sin \theta)^2 - 2g \left(\frac{u^2 \sin^2 \theta}{6g}\right) \]
Simplifying:
\[ v_y^2 = u^2 \sin^2 \theta - \frac{u^2 \sin^2 \theta}{3} = \frac{2u^2 \sin^2 \theta}{3} \]
Thus, the vertical component of velocity at one-third of maximum height is:
\[ v_y = \sqrt{\frac{2}{3}} u \sin \theta \]

### Step 3: Finding the total speed at one-third of maximum height
The total speed \( V \) at one-third of maximum height is given by:
\[ V = \sqrt{(u \cos \theta)^2 + \left(\sqrt{\frac{2}{3}} u \sin \theta\right)^2} \]
Calculating this:
\[ V = \sqrt{u^2 \cos^2 \theta + \frac{2}{3} u^2 \sin^2 \theta} = u \sqrt{\cos^2 \theta + \frac{2}{3} \sin^2 \theta} \]

### Step 4: Setting up the equation based on the problem statement
According to the problem, the speed at maximum height is \( \sqrt{\frac{5}{9}} \) times the speed at one-third of maximum height:
\[ u \cos \theta = \sqrt{\frac{5}{9}} \cdot u \sqrt{\cos^2 \theta + \frac{2}{3} \sin^2 \theta} \]
Dividing both sides by \( u \) (assuming \( u \neq 0 \)):
\[ \cos \theta = \sqrt{\frac{5}{9}} \sqrt{\cos^2 \theta + \frac{2}{3} \sin^2 \theta} \]

### Step 5: Squaring both sides
Squaring both sides gives:
\[ \cos^2 \theta = \frac{5}{9} \left(\cos^2 \theta + \frac{2}{3} \sin^2 \theta\right) \]
Expanding and rearranging:
\[ 9 \cos^2 \theta = 5 \cos^2 \theta + \frac{10}{3} \sin^2 \theta \]
\[ 4 \cos^2 \theta = \frac{10}{3} \sin^2 \theta \]
Multiplying through by 3:
\[ 12 \cos^2 \theta = 10 \sin^2 \theta \]
Dividing by \( \cos^2 \theta \):
\[ 12 = 10 \tan^2 \theta \]
Thus:
\[ \tan^2 \theta = \frac{12}{10} = \frac{6}{5} \]

### Step 6: Finding the angle \( \theta \)
Taking the square root:
\[ \tan \theta = \sqrt{\frac{6}{5}} \]
Thus:
\[ \theta = \tan^{-1} \left(\sqrt{\frac{6}{5}}\right) \]

### Step 7: Final calculation
Using a calculator, we find:
\[ \theta \approx 47.6^\circ \]

### Summary of Steps
1. Identify the components of the projectile's velocity.
2. Use the equations of motion to find the vertical component at one-third of maximum height.
3. Calculate the total speed at one-third of maximum height.
4. Set up the equation based on the given relationship between speeds.
5. Square both sides to eliminate the square root and simplify.
6. Solve for \( \tan \theta \) and find \( \theta \).