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Question

If a ball is thrown up at a velocity of 20 m/s from position ‘x‘ 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 B -20 m/s
Given:
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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