Question

The range of the projectile projected at an angle of 15° with hor izontal is 50 m. If the projectile is projected with same velocity at an angle of 45° with horizontal, then its range will be

• A. 50 m
• B. 100 m
• C. 80 m
• D. 120 m

Hint:

The range of a projectile is proportional to sin(2θ). Use the given range at 15 degrees to find the relationship between u2/g and the range. Then, calculate the range at 45 degrees using this relationship.

Answer 1

The Correct Option is B

Given:

• Angle of projection \( \theta_1 = 15^\circ \), range \( R_1 = 50 \, \text{m} \).
• Angle of projection \( \theta_2 = 45^\circ \).
• Initial velocity \( u \) remains the same.

Step 1: Formula for Range

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

Step 2: Relationship Between Ranges

For two angles \( \theta_1 = 15^\circ \) and \( \theta_2 = 45^\circ \), we have:

\[ \frac{R_2}{R_1} = \frac{\sin(2\theta_2)}{\sin(2\theta_1)} \] Substituting \( \theta_1 = 15^\circ \) and \( \theta_2 = 45^\circ \): \[ \frac{R_2}{50} = \frac{\sin(90^\circ)}{\sin(30^\circ)} \]

\(\sin(90^\circ) = 1\) and \(\sin(30^\circ) = 0.5\), so: \[ \frac{R_2}{50} = \frac{1}{0.5} = 2 \]

Therefore: \[ R_2 = 50 \times 2 = 100 \, \text{m} \]

Step 3: Final Answer

The range of the projectile when projected at \( 45^\circ \) is \( \boxed{100 \, \text{m}} \).

Answer 2

Given that,

Range R=50m

We know that,

For the first case

 50= \(\frac{(sin^2×15^{\circ})u^2}{g}\)
50 = \(\frac{u^2}{2g}\)
\(\frac{u^2}{g}=100.....(I)\)
For the second case

R= \(\frac{(sin^2×15^{\circ})u^2}{g}\)
R= \(\frac{u^2}{g}\)
R=100m

Hence, the range is 100 m