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Physics · MOTION IN A STRAIGHT LINE

हिन्दी

Match List-I with List-II for motion under gravity (ball thrown vertically upward with initial speed \( v_0 \)): List-I List-II A Time taken to reach peak height I \( v_0 \) B Maximum height reached above launch point II \( \frac{2 v_0}{g} \) C Total time of flight to return to launch level III \( \frac{v_0}{g} \) D Speed upon returning to launch level IV \( \frac{v_0^2}{2g} \) Choose the correct answer from the options given below:

Match List-I with List-II for motion under gravity (ball thrown vertically upward with initial speed \( v_0 \)):


List-I List-II
A Time taken to reach peak height I \( v_0 \)
B Maximum height reached above launch point II \( \frac{2 v_0}{g} \)
C Total time of flight to return to launch level III \( \frac{v_0}{g} \)
D Speed upon returning to launch level IV \( \frac{v_0^2}{2g} \)

Choose the correct answer from the options given below:

  1. A
    A-(III), B-(I), C-(II), D-(IV)
  2. B
    A-(IV), B-(III), C-(II), D-(I)
  3. C
    A-(II), B-(IV), C-(III), D-(I)
  4. D
    A-(III), B-(IV), C-(II), D-(I)
View Answer & Solution

Correct answer · Option D

A-(III), B-(IV), C-(II), D-(I)

Explanation

Time to peak \( t = v_0 / g \) (A-III). Maximum height \( h_{\max} = v_0^2 / (2g) \) (B-IV). Total flight time \( T = 2v_0 / g \) (C-II). Speed returning to launch level equals initial speed \( v_0 \) (D-I).
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