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. range of projectile
C. max height acquired
D. all of them

Explanation

To solve the problem, we need to analyze the motion of two projectiles thrown at different angles but with the same initial velocity and mass. The angles given are \(60^\circ\) and \(30^\circ\).

### Step-by-Step Solution:

1. **Identify the Angles**:
- Let \( \theta_1 = 60^\circ \) (first projectile)
- Let \( \theta_2 = 30^\circ \) (second projectile)

2. **Check for Complementary Angles**:
- Complementary angles are two angles that add up to \(90^\circ\).
- Here, \( \theta_1 + \theta_2 = 60^\circ + 30^\circ = 90^\circ \).
- Since the angles are complementary, we can use the properties of projectile motion.

3. **Understanding Projectile Motion**:
- The range \( R \) of a projectile is given by the formula:
\[ R = \frac{v^2 \sin(2\theta)}{g} \]
where \( v \) is the initial velocity, \( g \) is the acceleration due to gravity, and \( \theta \) is the angle of projection.

4. **Calculate the Range for Both Angles**:
- For \( \theta_1 = 60^\circ \):
\[ R_1 = \frac{v^2 \sin(120^\circ)}{g} \]
(since \( \sin(120^\circ) = \sin(180^\circ - 60^\circ) = \sin(60^\circ) \))
- For \( \theta_2 = 30^\circ \):
\[ R_2 = \frac{v^2 \sin(60^\circ)}{g} \]

5. **Compare the Ranges**:
- Since \( \sin(120^\circ) = \sin(60^\circ) \), we have:
\[ R_1 = R_2 \]
- Therefore, the ranges of both projectiles are the same.

6. **Conclusion**:
- The quantity that remains the same for both projectiles is the **range**.

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

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