Question

From the ground, a projectile is fired at an angle of $60^\circ$ to the horizontal with a speed of $20$ m/s. Take acceleration due to gravity as $10$ m/s$^2$. The horizontal range of the projectile is

• A. $10\sqrt{3}$ m
• B. $20$ m
• C. $20\sqrt{3}$ m
• D. $40\sqrt{3}$ m
• E. $400\sqrt{3}$ m

Hint:

Use identity $\sin(180^\circ - \theta) = \sin \theta$ to simplify projectile problems.

Answer 1

The Correct Option is C

Concept:
Range of a projectile: \[ R = \frac{u^2 \sin 2\theta}{g} \]

Step 1: Substitute values.
\[ u = 20, \quad \theta = 60^\circ, \quad g = 10 \] \[ R = \frac{20^2 \sin 120^\circ}{10} \]

Step 2: Use trigonometric identity.
\[ \sin 120^\circ = \sin 60^\circ = \frac{\sqrt{3}}{2} \]

Step 3: Calculate range.
\[ R = \frac{400 \times \frac{\sqrt{3}}{2}}{10} = \frac{200\sqrt{3}}{10} = 20\sqrt{3} \]