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Physics-(general)

NEET Physics (general) MCQ Question

Type: MCQ-numerical-Easy-Class 11

If a projectile is launched with an initial velocity v0v_0 at an angle θ0\theta_0 to the horizontal, what is the mathematical expression for its horizontal range RR?

A

R=v02sin⁡2θ0gR = \frac{v_0^2 \sin 2\theta_0}{g}

B

R=v02cos⁡2θ0gR = \frac{v_0^2 \cos 2\theta_0}{g}

C

R=v02tan⁡2θ0gR = \frac{v_0^2 \tan 2\theta_0}{g}

D

R=v02sec⁡2θ0gR = \frac{v_0^2 \sec 2\theta_0}{g}

Correct Answer

Option A

Detailed Explanation

The horizontal range of a projectile is given by R=v02sin⁡2θ0gR = \frac{v_0^2 \sin 2\theta_0}{g}, as shown in the NCERT context discussing the range equation for projectile motion.

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