Question

The range of projectile is maximum when the angle of projection is:

• A. 45$^\circ$
• B. 30$^\circ$
• C. 60$^\circ$
• D. 90$^\circ$

Hint:

For a projectile, the range is maximum at 45° because \( \sin(2\theta) \) is maximum at 90°, i.e., \( \theta = 45° \).

Answer 1

The Correct Option is A

The range \( R \) of a projectile launched with initial velocity \( u \) at an angle \( \theta \) with the horizontal is given by the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \] Where:
- \( u \) is the initial velocity,
- \( \theta \) is the angle of projection,
- \( g \) is the acceleration due to gravity.
To maximize \( R \), \( \sin(2\theta) \) must be maximized. Since the maximum value of \( \sin \) is 1, we get:
\[ \sin(2\theta) = 1 ⇒ 2\theta = 90^\circ ⇒ \theta = 45^\circ \]
Therefore, the range is maximum when the angle of projection is 45$^\circ$.