Question

If the range of a body projected with a velocity of 60 m/s is \( 180\sqrt{3} \) m, then the angle of projection of the body is (Acceleration due to gravity = 10 m/s\(^2\))

• A. 30° or 60°
• B. 37° or 53°
• C. 20° or 70°
• D. 15° or 75°

Hint:

Use the range formula \( R = \frac{u^2 \sin 2\theta}{g} \). Note that \( \sin 2\theta = \sin (180^\circ - 2\theta) \) gives two angles: \( \theta \) and \( 90^\circ - \theta \).

Answer 1

The Correct Option is A

The range of a projectile is given by: \[ R = \frac{u^2 \sin 2\theta}{g} \] Given \( u = 60 \, \text{m/s} \), \( R = 180\sqrt{3} \, \text{m} \), \( g = 10 \, \text{m/s}^2 \): \[ 180\sqrt{3} = \frac{60^2 \sin 2\theta}{10} \] \[ 180\sqrt{3} = \frac{3600 \sin 2\theta}{10} = 360 \sin 2\theta \] \[ \sin 2\theta = \frac{180\sqrt{3}}{360} = \frac{\sqrt{3}}{2} \] \[ 2\theta = 60^\circ \quad \text{or} \quad 2\theta = 180^\circ - 60^\circ = 120^\circ \] \[ \theta = 30^\circ \quad \text{or} \quad \theta = 60^\circ \] Option (1) is correct. Options (2), (3), and (4) do not satisfy \( \sin 2\theta = \frac{\sqrt{3}}{2} \).