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Physics · MOTION IN A PLANE
हिन्दीMatch List-I with List-II: List-I List-II A Maximum height (\( h_m \)) I \( \frac{v_0^2 \sin 2\theta_0}{g} \) B Time of flight (\( T_f \)) II \( \frac{v_0^2 \sin^2\theta_0}{2g} \) C Horizontal range (\( R \)) III \( \frac{2v_0 \sin\theta_0}{g} \) D Equation of trajectory IV \( x \tan\theta_0 - \frac{g x^2}{2 v_0^2 \cos^2\theta_0} \) Choose the correct answer from the options given below:
Match List-I with List-II:
| List-I | List-II | ||
|---|---|---|---|
| A | Maximum height (\( h_m \)) | I | \( \frac{v_0^2 \sin 2\theta_0}{g} \) |
| B | Time of flight (\( T_f \)) | II | \( \frac{v_0^2 \sin^2\theta_0}{2g} \) |
| C | Horizontal range (\( R \)) | III | \( \frac{2v_0 \sin\theta_0}{g} \) |
| D | Equation of trajectory | IV | \( x \tan\theta_0 - \frac{g x^2}{2 v_0^2 \cos^2\theta_0} \) |
Choose the correct answer from the options given below:
Correct answer · Option 3
A-II, B-III, C-I, D-IV
Explanation
Standard kinematic expressions for projectile motion give maximum height \( h_m = \frac{v_0^2 \sin^2\theta_0}{2g} \), time of flight \( T_f = \frac{2v_0 \sin\theta_0}{g} \), range \( R = \frac{v_0^2 \sin 2\theta_0}{g} \), and trajectory \( y = x \tan\theta_0 - \frac{g x^2}{2 v_0^2 \cos^2\theta_0} \).
