BITSAT 2021 Paper

Let the velocity of the ball just when it leaves the hand is u then applying, v2−u2=2as for upward journey
⇒ −u2=2(−10)×2
⇒u2=40
Again applying v2−u2=2as for the upward journey of the ball, when the ball is in the hands of the thrower,
v2−u2=2as
⇒ 40−0=2(a)0.2
⇒a=100m∕s2
∴ F=ma=0.2×100=20N
⇒ N−mg=20
⇒N=20+2=22N