Question

When a ball A is thrown in vertically upward direction with velocity 20 ms-1, exactly at the same time another ball B is allowed to fall freely from height h. The magnitude of velocity of A with respect to B at time 2 sec is ... (g = 10 m/s2.) (A) 40 (B) 10 (C) 20 (D) 30​

Answer 1

Answer:

5 seconds

Explanation:

Time to reach maximum height can be obtained from v = u + at

0=20+(−10)t

t = 2 seconds

s=ut+0.5at2 = 20(2)+0.5(−10)(2)2 = 20m

Thus, total distance for maximum height is 45 m

s=ut+0.5at2

45=0+0.5(10)(t')2

t′  = 3s

Total time= 3+2= 5s

Answer 2

Answer:

The correct option is (C)20 m/s

Explanation:

• For the ball, thrown upward: Initial velocity, u = 20 m/s

       Acceleration = g = - 10 m/s² (in upward direction g is negative)

        time = t = 2 s.

       From 1st equation of motion, v = u + gt = 20 - 10(2) = 20 - 20 = 0 m/s.

After 2 sec the ball's velocity becomes 0.

• For the ball, falling downward: Initial velocity, u = 0 m/s Acceleration = g = 10 m/s² ,  time = t = 2 s
• From 1st equation of motion, v = u + gt = 0 + 10(2) = 20 m/s.
• After 2 sec the ball's velocity becomes 20 m/s.

• Relative velocity: As ball A and B are moving in opposite directions, so velocity of A with respect to B = velocity of A - (-velocity of B)

                                                                = 0 + 20 = 20 m/s.

The magnitude of velocity of A with respect to B at time 2 sec is 20 m/s.

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