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Physics · JEE | MOTION IN A PLANE

हिन्दी

For a given projection velocity \( v_0 \), what is the formula for the maximum horizontal range \( R_m \) of a projectile?

  1. A
    \( R_m = \frac{2v_0^2}{g} \)
  2. B
    \( R_m = \frac{v_0^2}{g} \)
  3. C
    \( R_m = \frac{v_0^2 \sin^2 \theta_0}{g} \)
  4. D
    \( R_m = \frac{v_0 \sin 2\theta_0}{2g} \)
View Answer & Solution

Correct answer · Option B

\( R_m = \frac{v_0^2}{g} \)

Explanation

Range is maximum when \( \sin 2\theta_0 = 1 \) (i.e. \( \theta_0 = 45^\circ \)), giving \( R_m = v_0^2/g \).
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