AP EAPCET 05-Jul-2022 Shift 2 Paper

The given situation is shown in the figureEF‌=EB−FB
‌=EB−GA
‌=8.75−3.75=5m
∴In△EFG
‌tan‌30∘=‌

EF

GF

‌⇒‌‌‌

1

√3

=‌

5

GF

‌⇒‌‌GF=5√3m
Total time taken by the sphere to reach the ground is t, then ‌h=‌

1

2

gt2
‌8.75=‌

1

2

×10×t2⇒‌

17.50

10

=t2
⇒‌‌t2‌=1.75
⇒‌‌t=√1.75 s=1.32 s
If t1 be the time taken to cover a distance of EF, then,
5=‌

1

2

gt12
⇒‌‌t1=1 s
Time taken to cover distance FB
t′=t−t1=1.32−1=0.32 s
If v be the velocity of sphere at point G, then gravitational potential energy at point E
= Gravitational potential energy at point G
mgh+‌

1

2

Iω2+‌

1

2

mv2
‌⇒‌‌5g=‌

7

10

v2
Horizontal component of velocity at point G,
vx′‌=v‌cos‌30∘
Range, OA‌=vx′×t′
‌=5×√‌

15

7

×0.32=1.5√‌

15

7

m
∴‌‌OB‌=AB+OA
‌=GF+OA
‌=(5√3+1.5√‌

15

7

)m
‌=10.1m≃10m