JEE Main 26-Aug-2021 Shift 1 Solved Paper

Given, m1=m2=m
r1=r2=r=5‌cm=5×10−2m
u1=u2=35m∕s
t1=0
and t2=3s
When both ball will collide,
(S1)vertical=(S2)vertical
Let after time t both ball will collide.
(S1)vertical=u1t−

1

2

gt2
(S2)vertical=u2(t−3)−

1

2

g(t−3)2
From Eq. (i), we have
⇒u1t−

1

2

gt2=u2(t−3)−

1

2

g(t−3)2
⇒35t−

1

2

×10×t2 =35(t−3)−

1

2

×10(t−3)2
⇒35t−35(t−3)=5t2−5(t−3)2
⇒35(t−t+3)=5[t2−(t−3)2]
35×3=5(t+t−3)(t−t+3)
⇒

105

5

=(2t−3)(3)
⇒

21

3

=2t−3
⇒7=2t−3
⇒2t=10
⇒t=5s
So, the height at which ball will collide is
(S1)vertical=u1t−

1

2

gt2
=35×5−

1

2

×10×(5)2
=175−125
=50m