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Question

# A ball is projected vertically up with a velocity of 20 ms−1. Its velocity at the height of 15 m is (Take g=10 ms−2)

A
10 ms1
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B
15 ms1
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C
10 ms1
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D
Both a and c
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Solution

## The correct option is D Both a and cGiven, u=20 ms−1,s=15 m We know that v2−u2=2as v2−(20)2=−2×10×15 v2=100⇒v=±10 ms−1 Hence, the ball will cross the height two times, first time when it goes up and second when it comes down. While going up its velocity at the height of 15 m will be 10 m/s and while coming down the velocity will be −10 m/s.

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