Kinematics — Easy JEE physics MCQ
A projectile is thrown with velocity $20$ m/s at an angle of $30°$ with the horizontal. The horizontal range of the projectile is (Take $g = 10$ m/s²):
- A. $20$ m
- B. $20\sqrt{3}$ m
- C. $40$ m
- D. $40\sqrt{3}$ m
Solution
The range formula for projectile motion:
$$R = \frac{u^2 \sin 2\theta}{g}$$
Given: $u = 20$ m/s, $\theta = 30°$, $g = 10$ m/s²
$$R = \frac{(20)^2 \sin 60°}{10} = \frac{400 \times \frac{\sqrt{3}}{2}}{10} = \frac{200\sqrt{3}}{10} = 20\sqrt{3} \text{ m}$$