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

Options

A. `(a)alpha + beta = pi/2`
B. `(b)R = 4(sqrt(h_1h_2))`
C. `(c)t_1/t_2 = tan alpha`
D. `(d)tan alpha = (sqrt (h_1/h_2))`

Explanation

To solve the problem, we need to analyze the motion of two projectiles launched from the same point with the same speed \( u \) but at different angles \( \alpha \) and \( \beta \). Since they strike the ground at the same point, we can derive relationships between their maximum heights, times of flight, and the range.

### Step-by-Step Solution:

1. **Understanding the Range Condition**:
Since both projectiles strike the ground at the same point, their ranges must be equal. The range \( R \) for a projectile launched at an angle \( \theta \) is given by:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
Therefore, for angles \( \alpha \) and \( \beta \):
\[ R = \frac{u^2 \sin(2\alpha)}{g} = \frac{u^2 \sin(2\beta)}{g} \]
This implies:
\[ \sin(2\alpha) = \sin(2\beta) \]
Since \( \alpha \) and \( \beta \) are complementary angles, we have:
\[ \alpha + \beta = \frac{\pi}{2} \]

2. **Finding the Time of Flight**:
The time of flight \( t \) for a projectile is given by:
\[ t = \frac{2u \sin(\theta)}{g} \]
Thus, for the two angles:
\[ t_1 = \frac{2u \sin(\alpha)}{g}, \quad t_2 = \frac{2u \sin(\beta)}{g} \]
Taking the ratio of the times of flight:
\[ \frac{t_1}{t_2} = \frac{\sin(\alpha)}{\sin(\beta)} \]
Using the complementary angle identity, \( \sin(\beta) = \cos(\alpha) \):
\[ \frac{t_1}{t_2} = \frac{\sin(\alpha)}{\cos(\alpha)} = \tan(\alpha) \]

3. **Finding the Maximum Heights**:
The maximum height \( h \) for a projectile is given by:
\[ h = \frac{u^2 \sin^2(\theta)}{2g} \]
Therefore, for the two projectiles:
\[ h_1 = \frac{u^2 \sin^2(\alpha)}{2g}, \quad h_2 = \frac{u^2 \sin^2(\beta)}{2g} \]
Taking the ratio of the maximum heights:
\[ \frac{h_1}{h_2} = \frac{\sin^2(\alpha)}{\sin^2(\beta)} = \frac{\sin^2(\alpha)}{\cos^2(\alpha)} = \tan^2(\alpha) \]

4. **Verifying the Range in Terms of Heights**:
The range can also be expressed in terms of the maximum heights:
\[ R = 4 \sqrt{h_1 h_2} \]
This is derived from the relationships established earlier.

### Summary of Relationships:
- \( \alpha + \beta = \frac{\pi}{2} \)
- \( \frac{t_1}{t_2} = \tan(\alpha) \)
- \( \frac{h_1}{h_2} = \tan^2(\alpha) \)
- \( R = 4 \sqrt{h_1 h_2} \)