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.

Take the upward direction to be positive.

- 9.8 m/s
- -20 m/s
- 0 m/s
- 20 m/s

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

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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