Two projectiles of same mass and with same velocity are thrown at an angle `60^(@)` and `30^(@)` with the horizontal, then which quantity will remain same:-

Options

A. Time of flight
B. Horizontal range of projectile
C. Max height acquired
D. All of them

Explanation

To solve the problem, we need to analyze the projectile motion of two projectiles launched at angles of \(60^\circ\) and \(30^\circ\) with the same initial velocity. We want to determine which quantity remains the same for both projectiles.

### Step 1: Understand the angles and their relationship
We have two angles:
- \(\theta_1 = 60^\circ\)
- \(\theta_2 = 30^\circ\)

These angles are complementary, meaning:
\[ \theta_1 + \theta_2 = 90^\circ \]

### Step 2: Use the range formula for projectile motion
The horizontal range \(R\) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
where:
- \(u\) is the initial velocity,
- \(g\) is the acceleration due to gravity,
- \(\theta\) is the angle of projection.

### Step 3: Calculate the range for both angles
1. **For \(\theta_1 = 60^\circ\)**:
\[ R_1 = \frac{u^2 \sin(2 \times 60^\circ)}{g} = \frac{u^2 \sin(120^\circ)}{g} \]
Since \(\sin(120^\circ) = \sin(180^\circ - 60^\circ) = \sin(60^\circ) = \frac{\sqrt{3}}{2}\):
\[ R_1 = \frac{u^2 \cdot \frac{\sqrt{3}}{2}}{g} = \frac{\sqrt{3} u^2}{2g} \]

2. **For \(\theta_2 = 30^\circ\)**:
\[ R_2 = \frac{u^2 \sin(2 \times 30^\circ)}{g} = \frac{u^2 \sin(60^\circ)}{g} \]
Since \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\):
\[ R_2 = \frac{u^2 \cdot \frac{\sqrt{3}}{2}}{g} = \frac{\sqrt{3} u^2}{2g} \]

### Step 4: Compare the ranges
From the calculations:
\[ R_1 = R_2 = \frac{\sqrt{3} u^2}{2g} \]
Thus, the horizontal range for both projectiles is the same.

### Step 5: Conclusion
The quantity that remains the same for both projectiles is the **horizontal range**.

### Final Answer:
The horizontal range of the projectiles will remain the same.

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