Two particles projected form the same point with same speed u at angles of projection `alpha and beta` strike the horizontal ground at the same point. If `h_1 and h_2` are the maximum heights attained by the projectile, R is the range for both and `t_1 and t_2` are their times of flights, respectively, then incorrect option is :

Options

A. `alpha + beta = (pi)/(2)`
B. `R = 4 sqrt(h_(1)h_(2))`
C. `tan alpha = (t_(1))/(t_(2)) = sqrt(h_(1)h_(2))`
D. None of the above

Explanation

To solve the problem, we need to analyze the motion of two projectiles launched from the same point with the same speed \( u \) at angles \( \alpha \) and \( \beta \) such that they strike the ground at the same point. We will derive the expressions for maximum heights, times of flight, and ranges, and then identify the incorrect option among the given statements.

### Step-by-Step Solution:

1. **Understanding the Condition**:
Since both projectiles strike the ground at the same point, the angles \( \alpha \) and \( \beta \) must be complementary. This means:
\[ \alpha + \beta = 90^\circ \]

2. **Maximum Heights**:
The maximum height \( h_1 \) for the projectile launched at angle \( \alpha \) is given by:
\[ h_1 = \frac{u^2 \sin^2 \alpha}{2g} \]
For the projectile launched at angle \( \beta \):
\[ h_2 = \frac{u^2 \sin^2 \beta}{2g} \]
Since \( \beta = 90^\circ - \alpha \), we can use the identity \( \sin(90^\circ - \alpha) = \cos \alpha \):
\[ h_2 = \frac{u^2 \cos^2 \alpha}{2g} \]

3. **Times of Flight**:
The time of flight \( t_1 \) for the projectile at angle \( \alpha \) is:
\[ t_1 = \frac{2u \sin \alpha}{g} \]
For the projectile at angle \( \beta \):
\[ t_2 = \frac{2u \sin \beta}{g} = \frac{2u \cos \alpha}{g} \]

4. **Range**:
The range \( R \) for both projectiles can be calculated using the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
For angle \( \alpha \):
\[ R = \frac{u^2 \sin(2\alpha)}{g} \]
Since \( \beta = 90^\circ - \alpha \):
\[ R = \frac{u^2 \sin(2(90^\circ - \alpha))}{g} = \frac{u^2 \sin(180^\circ - 2\alpha)}{g} = \frac{u^2 \sin(2\alpha)}{g} \]
Thus, both projectiles have the same range \( R \).

5. **Comparing Relationships**:
- We need to check the relationships given in the options:
1. \( R = 4 \sqrt{h_1 h_2} \)
2. \( \frac{t_1}{t_2} = \tan \alpha \)
3. \( \frac{t_1}{t_2} \neq \sqrt{h_1 h_2} \)

- **Calculating \( 4 \sqrt{h_1 h_2} \)**:
\[ \sqrt{h_1 h_2} = \sqrt{\left(\frac{u^2 \sin^2 \alpha}{2g}\right) \left(\frac{u^2 \cos^2 \alpha}{2g}\right)} = \frac{u^2 \sin \alpha \cos \alpha}{2g} \]
Therefore:
\[ 4 \sqrt{h_1 h_2} = 4 \cdot \frac{u^2 \sin \alpha \cos \alpha}{2g} = \frac{2u^2 \sin(2\alpha)}{g} = R \]
This statement is correct.

- **Calculating \( \frac{t_1}{t_2} \)**:
\[ \frac{t_1}{t_2} = \frac{2u \sin \alpha / g}{2u \cos \alpha / g} = \frac{\sin \alpha}{\cos \alpha} = \tan \alpha \]
This statement is also correct.

- **Checking \( \frac{t_1}{t_2} \) against \( \sqrt{h_1 h_2} \)**:
\[ \sqrt{h_1 h_2} = \frac{u^2 \sin \alpha \cos \alpha}{2g} \]
We already found \( \frac{t_1}{t_2} = \tan \alpha \), which is not equal to \( \sqrt{h_1 h_2} \) since \( \tan \alpha \) is not equal to \( \frac{u^2 \sin \alpha \cos \alpha}{2g} \).

### Conclusion:
The incorrect option is:
\[ \text{Option 3: } \frac{t_1}{t_2} \neq \sqrt{h_1 h_2} \]