Two projectiles `A` and `B` are projected with same speed at an angle `30^(@)` and `60^(@)` to the horizontal, then which of the following is not valid where `T` is total time of flight, `H` is max height and `R` is horizontal range.

Options

A. `H_(A) = 3H_(B)`
B. `T_(B) = sqrt(3)T_(A)`
C. `R_(A) = R_(B)`
D. `H_(B)-3H_(A)`

Answer

To solve the problem, we need to analyze the projectile motion of two projectiles `A` and `B` launched at angles of `30°` and `60°` respectively, with the same initial speed `u`. We will calculate the total time of flight (`T`), maximum height (`H`), and horizontal range (`R`) for both projectiles and determine which statement is not valid.

### Step 1: Calculate Time of Flight for Projectile A (30°)
The formula for the time of flight \( T \) for a projectile is given by:
\[ T = \frac{2u \sin \theta}{g} \]
For projectile A:
\[ T_A = \frac{2u \sin 30°}{g} = \frac{2u \cdot \frac{1}{2}}{g} = \frac{u}{g} \]

### Step 2: Calculate Time of Flight for Projectile B (60°)
For projectile B:
\[ T_B = \frac{2u \sin 60°}{g} = \frac{2u \cdot \frac{\sqrt{3}}{2}}{g} = \frac{\sqrt{3}u}{g} \]

### Step 3: Calculate Maximum Height for Projectile A (30°)
The formula for maximum height \( H \) is given by:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
For projectile A:
\[ H_A = \frac{u^2 \sin^2 30°}{2g} = \frac{u^2 \cdot \left(\frac{1}{2}\right)^2}{2g} = \frac{u^2 \cdot \frac{1}{4}}{2g} = \frac{u^2}{8g} \]

### Step 4: Calculate Maximum Height for Projectile B (60°)
For projectile B:
\[ H_B = \frac{u^2 \sin^2 60°}{2g} = \frac{u^2 \cdot \left(\frac{\sqrt{3}}{2}\right)^2}{2g} = \frac{u^2 \cdot \frac{3}{4}}{2g} = \frac{3u^2}{8g} \]

### Step 5: Calculate Horizontal Range for Projectile A (30°)
The formula for horizontal range \( R \) is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
For projectile A:
\[ R_A = \frac{u^2 \sin 60°}{g} = \frac{u^2 \cdot \frac{\sqrt{3}}{2}}{g} = \frac{\sqrt{3}u^2}{2g} \]

### Step 6: Calculate Horizontal Range for Projectile B (60°)
For projectile B:
\[ R_B = \frac{u^2 \sin 120°}{g} = \frac{u^2 \cdot \frac{\sqrt{3}}{2}}{g} = \frac{\sqrt{3}u^2}{2g} \]

### Summary of Results
- Time of Flight:
- \( T_A = \frac{u}{g} \)
- \( T_B = \frac{\sqrt{3}u}{g} \)

- Maximum Height:
- \( H_A = \frac{u^2}{8g} \)
- \( H_B = \frac{3u^2}{8g} \)

- Horizontal Range:
- \( R_A = \frac{\sqrt{3}u^2}{2g} \)
- \( R_B = \frac{\sqrt{3}u^2}{2g} \)

### Validity Check of Statements
1. **\( H_A = 3H_B \)**: This is incorrect because \( H_A = \frac{u^2}{8g} \) and \( H_B = \frac{3u^2}{8g} \) implies \( H_A \neq 3H_B \).
2. **\( T_B = \sqrt{3} T_A \)**: This is correct because \( T_B = \frac{\sqrt{3}u}{g} \) and \( T_A = \frac{u}{g} \).
3. **\( R_A = R_B \)**: This is correct since both ranges are equal.
4. **\( H_A - 3H_B = 0 \)**: This is incorrect since \( H_A - 3H_B \) does not equal zero.

### Conclusion
The statement that is **not valid** is **\( H_A = 3H_B \)**.