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
• B. 9.8 m/s
• C. 20 m/s
• D. -20 m/s

Answer 1

The correct option is D -20 m/s

Given:
Upward direction is positive
Initial and final positions are the same. So, displacement is:
$s=0m$
Body is under free fall. So, acceleration is:
$a=-g=-9.8m/s^2$
Initial velocity, $u=20m/s$

Using equation of motion,
$v^2=u^2+2as$
$v^2={20}^2+2\times (-9.8)\times \left(0\right)$
$v^2={20}^2$
$v=\pm 20m/s$

Since the ball is going down, the velocity will be negative.
$v=-20m/s$