CGPET 2019 Solved Paper

First of all we divide the journey of ball in two parts, upward and downward.
Case I For upward journey,
From work energy theorem,
Work done, W=Kf−Ki=ΔK
⇒−Fh−mgh=−

1

2

mv2−0
⇒Fh+mgh=

1

2

mv2 .......(i)
Since, F is a resistance force.
Case II For downward journey,
From work energy theorem
W=Kf−Ki=ΔK
⇒−Fh+mgh=

1

2

mv12 .......(ii)
(here, v1= final velocity )
Divide (i) and (ii), we get

(Fh+mgh)

(−Fh+mgh)

=

1 2 mv2

1 2 mv12

⇒

h(F+mg)

h(−F+mg)

=

v2

v12

⇒ v12=v2

(mg−F)

(mg+F)

Hence, the velocity with which the ball returns to the thrower is
v1=v√

(mg−F)

(mg+F)