AMU Engineering 2020 solved paper Solved Paper

Given, ‌‌u=60m∕s,R=180m
∴‌‌R=‌

u2

g

‌sin‌2‌θ
‌‌180=‌

(60)2

10

‌sin‌2‌θ
⇒‌‌sin‌2‌θ=‌

1

2

⇒‌‌θ=15∘
As there are always two directions of projections θ and(90∘−θ) for the given range, hence two directions of projection, i.e. 15∘ and 90∘−15∘, i.e. 15∘ and 75∘
Let T1 and T2 be the times of flight in the two cases, where
T1=‌

2u‌sin‌15∘

g

T2=‌

2u‌sin‌75∘

g

The man is in danger for a time
T2−T1‌‌=‌

2u

g

(sin‌75∘−sin‌15∘)
‌‌=‌

2u

g

(2‌cos‌45∘‌sin‌30∘)
[using identity,
sin‌C−sin‌D=2‌cos‌

C+D

2

‌sin‌

C−D

2

]
‌‌=‌

2×60

10

×2×‌

1

√2

×‌

1

2

=6√2s