Two particles are projected in air with speed `v_(0)` at angles `theta_(1)` and `theta_(2)` (both acute) to the horizontal,respectively.If the height reached by the first particle greater than that of the second,then thick the right choices

Options

A. Angle of projection `q_(1) gt q_(2)`
B. Time of flight `T_(1) gt T_(2)`
C. Horizontal range `R_(1) gt R_(2)`
D. Total energy `U_(1) gt U_(2)`

Explanation

To solve the problem step by step, we need to analyze the motion of the two particles projected at different angles but with the same initial speed.

### Step 1: Understanding the maximum height formula
The maximum height \( h \) reached by a projectile is given by the formula:
\[ h = \frac{u^2 \sin^2 \theta}{2g} \]
where \( u \) is the initial speed, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.

### Step 2: Setting up the equations for both particles
Let:
- \( h_1 \) be the height reached by the first particle (angle \( \theta_1 \))
- \( h_2 \) be the height reached by the second particle (angle \( \theta_2 \))

From the formula, we have:
\[ h_1 = \frac{v_0^2 \sin^2 \theta_1}{2g} \]
\[ h_2 = \frac{v_0^2 \sin^2 \theta_2}{2g} \]

### Step 3: Comparing the heights
According to the problem, \( h_1 > h_2 \). Thus:
\[ \frac{v_0^2 \sin^2 \theta_1}{2g} > \frac{v_0^2 \sin^2 \theta_2}{2g} \]
Since \( v_0 \) and \( g \) are the same for both particles, we can simplify this to:
\[ \sin^2 \theta_1 > \sin^2 \theta_2 \]

### Step 4: Analyzing the sine function
Since both angles \( \theta_1 \) and \( \theta_2 \) are acute (between 0 and 90 degrees), the sine function is increasing in this interval. Therefore:
\[ \sin \theta_1 > \sin \theta_2 \implies \theta_1 > \theta_2 \]
This confirms that the angle of projection for the first particle is greater than that of the second.

### Step 5: Time of flight comparison
The time of flight \( T \) for a projectile is given by:
\[ T = \frac{2u \sin \theta}{g} \]
For both particles:
\[ T_1 = \frac{2v_0 \sin \theta_1}{g} \]
\[ T_2 = \frac{2v_0 \sin \theta_2}{g} \]
Since \( v_0 \) and \( g \) are the same, we can compare:
\[ T_1 > T_2 \quad \text{(because } \sin \theta_1 > \sin \theta_2\text{)} \]

### Step 6: Horizontal range comparison
The horizontal range \( R \) of a projectile is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
For both particles:
\[ R_1 = \frac{v_0^2 \sin 2\theta_1}{g} \]
\[ R_2 = \frac{v_0^2 \sin 2\theta_2}{g} \]
Since \( v_0 \) and \( g \) are the same, we can compare:
\[ R_1 > R_2 \quad \text{(because } 2\theta_1 > 2\theta_2 \text{ implies } \sin 2\theta_1 > \sin 2\theta_2\text{)} \]

### Step 7: Total mechanical energy comparison
The total mechanical energy \( E \) of a projectile is given by:
\[ E = \text{Potential Energy} + \text{Kinetic Energy} \]
Since both particles are projected with the same initial speed \( v_0 \), their total mechanical energy depends on their mass. If \( m_1 \) and \( m_2 \) are the masses of the first and second particles, respectively, we cannot definitively compare their energies without knowing their masses.

### Conclusion
From the analysis:
1. \( \theta_1 > \theta_2 \) (Angle of projection)
2. \( T_1 > T_2 \) (Time of flight)
3. \( R_1 > R_2 \) (Horizontal range)
4. Total energy cannot be compared without knowing the masses.

Thus, the correct choices are options 1, 2, and 3.