Question

# If a ball is thrown up at a velocity of 20 m/s from position â€˜xâ€˜ as shown in the figure. What will be the velocity of the ball upon reaching back to the same position? Take the upward direction to be positive.

A
0 m/s
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B
9.8 m/s
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C
20 m/s
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D
-20 m/s
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Solution

## The correct option is D -20 m/sGiven: Upward direction is positive Initial and final positions are the same. So, displacement is: s=0 m Body is under free fall. So, acceleration is: a=−g=−9.8 m/s2 Initial velocity, u=20 m/s Using equation of motion, v2=u2+2as v2=202+2×(−9.8)×(0) v2=202 v=±20 m/s Since the ball is going down, the velocity will be negative. v=−20 m/s

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