Question

There are two projectiles thrown at angles \( \theta_1 \) & \( \theta_2 \) such that their ranges are same. Their speeds of projection are also same and time periods are 10 sec and 5 sec respectively. Find the range.

• A. \(250\ \text{m}\)
• B. \(300\ \text{m}\)
• C. \(650\ \text{m}\)
• D. \(100\ \text{m}\)

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For two projectiles projected with the same speed to have the same horizontal range, their angles of projection must be complementary.
That is, if one angle is $\theta$, the other must be $90^\circ - \theta$.
Step 2: Key Formula or Approach:
The time of flight for a projectile is given by $T = \frac{2u\sin\theta}{g}$.
The horizontal range is given by $R = \frac{u^2 \sin 2\theta}{g} = \frac{2u^2 \sin\theta \cos\theta}{g}$.
Relation between Range and Times of Flight is $R = \frac{1}{2} g T_1 T_2$.
Step 3: Detailed Explanation:
Given $T_1 = 10$ sec and $T_2 = 5$ sec.
Using the individual components for time of flight:
\[ T_1 = \frac{2u\sin\theta}{g} = 10 \implies u\sin\theta = 50 \]
\[ T_2 = \frac{2u\cos\theta}{g} = 5 \implies u\cos\theta = 25 \]
Now, calculate the range using the derived components:
\[ R = \frac{2(u\sin\theta)(u\cos\theta)}{g} \]
\[ R = \frac{2(50)(25)}{10} = 250 \text{ m} \]
Step 4: Final Answer:
The range of the projectile is 250 m.