MarksRiser
MarksRiser
Physics-(general)

NEET Physics (general) MCQ Question

Type: MCQ-numerical-Medium-Class 11

A projectile is launched with an initial velocity vov_o at an angle θ0\theta_0 with the horizontal. If the initial speed vo=20 m/sv_o = 20 \text{ m/s} and the angle θ0=30∘\theta_0 = 30^\circ, what is the horizontal range RR of the projectile?

A

34.64 m

B

30.00 m

C

25.98 m

D

40.00 m

Correct Answer

Option A

Detailed Explanation

The horizontal range RR is given by R=vo2sin⁡2θ0gR = \frac{v_o^2 \sin 2\theta_0}{g}. Substituting vo=20 m/sv_o = 20 \text{ m/s}, θ0=30∘\theta_0 = 30^\circ, and g=9.8 m/s2g = 9.8 \text{ m/s}^2, we get R=202sin⁡60∘9.8=34.64 mR = \frac{20^2 \sin 60^\circ}{9.8} = 34.64 \text{ m}.

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