Question

The range of a projectile projected at an angle of \( 15^\circ \) with the horizontal is \( 50 \) m. If the projectile is projected with the same velocity at an angle of \( 45^\circ \), then its range will be:

• A. \( 50 \) m
• B. \( 50\sqrt{2} \) m
• C. \( 100 \) m
• D. \( 100\sqrt{2} \) m

Hint:

For a given velocity, the maximum range occurs at \( 45^\circ \). The range is proportional to \( \sin(2\theta) \), which helps in comparative range calculations.

Answer 1

The Correct Option is C

Step 1: {Use the range formula}
\[ R = \frac{v^2 \sin 2\theta}{g} \] Since \( R \propto \sin(2\theta) \), we use the ratio: \[ \frac{R_1}{R_2} = \frac{\sin(2\theta_1)}{\sin(2\theta_2)} \] Step 2: {Substituting values}
\[ \frac{50}{R_2} = \frac{\sin(30^\circ)}{\sin(90^\circ)} \] \[ = \frac{\frac{1}{2}}{1} = \frac{1}{2} \] \[ R_2 = 100 { m} \] Thus, the correct answer is (C) 100 m.

Answer 2

Given:
- A projectile is projected at an angle of \( 15^\circ \) with a certain velocity and has a range of \( 50 \) m.
- We are to find the new range when the same projectile is launched with the same velocity but at an angle of \( 45^\circ \).

Step 1: Use the range formula
The range of a projectile is given by the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
Where:
- \( u \) = initial velocity
- \( \theta \) = angle of projection
- \( g \) = acceleration due to gravity

Let \( R_1 \) be the initial range at \( 15^\circ \), and \( R_2 \) be the new range at \( 45^\circ \).
So:
\[ R_1 = \frac{u^2 \sin(30^\circ)}{g}, \quad R_2 = \frac{u^2 \sin(90^\circ)}{g} \]
We know:
- \( \sin(30^\circ) = \frac{1}{2} \)
- \( \sin(90^\circ) = 1 \)

So:
\[ \frac{R_2}{R_1} = \frac{\sin(90^\circ)}{\sin(30^\circ)} = \frac{1}{1/2} = 2 \Rightarrow R_2 = 2 \cdot R_1 = 2 \cdot 50 = 100 \text{ m} \]

Final Answer:
The new range of the projectile is 100 m.