‌⇒‌‌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=√1.75s=1.32s If t1 be the time taken to cover a distance of EF, then, 5=‌
1
2
gt12 ⇒‌‌t1=1s Time taken to cover distance FB t′=t−t1=1.32−1=0.32s 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
mg×8.75=mg×3.75+‌
1
2
⋅‌
2
5
mR2(‌
v
R
)2+‌
1
2
mv2 ⇒8.75g=3.75g+‌
1
5
×v2+‌
1
2
v2
‌⇒‌‌5g=‌
7
10
v2
‌⇒‌‌v=√‌
5g×10
7
=√‌
5×10×10
7
=10√‌
5
7
m/s
Horizontal component of velocity at point G, vx′‌=v‌cos‌30∘
‌=10√‌
5
7
‌cos‌30∘=10√‌
5
7
×‌
√3
2
=5√‌
15
7
m∕s
Range, OA‌=vx′×t′ ‌=5×√‌
15
7
×0.32=1.5√‌
15
7
m ∴‌‌OB‌=AB+OA ‌=GF+OA ‌=(5√3+1.5√‌