Explanation

(d) The situation is shown in figure. At initial time, the ball is at $P,$ then under the action of a force (exerted by hand) from P to A and then from A to B, let acceleration of ball during PA $is\ a\ {ms}^{- 2}\lbrack$assumed to be constant] in upward direction and velocity of ball at A is $v\ {ms}^{- 1}.$
Then for PA , $v^{2} = 0^{2} + 2a \times 0.2$
For AB , $0 = v^{2} - 2 \times g \times 2$
$\Longrightarrow \ \ \ \ v^{2} = 2g \times 2$
From above equation,
$a = 10g = 100\ {ms}^{- 2}$
Then for PA, FBD of ball is
$F - mg = ma\ \lbrack F$ is the force exerted by hand on ball]
$\Longrightarrow \ \ F = m(g + a) = 0.2\ (11g)$
$= 22\ N$
[image:5_image175.emf]
[image:5_image176.emf]