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Question

A ball is dropped from a building of height 45m. Simultaneously another ball is thrown up with a speed 40m/s. The relative speed of the balls as a function of time is

A
20ms1
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B
40ms1
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C
30ms1
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D
0ms1
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Solution

The correct option is B 40ms1
According to the problem, for the ball dropped from the building, u1=0 and u2=40m/s

Velocity of the ball after time t,

v1=u1gt

v1=gt

And for another ball which is thrown upward,

u2=40m/s

Velocity of the ball after time t,

v2=u2gt=(40gt)

Therefore,

Relative velocity of one ball with respect to another ball is,

v12=v1v2=gt[40gt]

v12=v1v2=gt+40+gt=40m/s

1474441_1433131_ans_4a0a19314d7f483881dd56b9f827b60c.png

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