TS EAMCET 18-July-2022 Shift 1 Solved Paper

The given situation is shown below

Since, time of flight, T=‌

2usin‌θ

g

‌=‌

2×40×sin‌30∘

10

‌=‌

2×40×1

10×2

=4 s
∴ In 2 s , it will be at the topmost point, where it will hit the hillside.
After 2 s, the maximum height (Hmax)
AB‌=Hmax=‌

u2sin‌2θ

2g

‌=‌

402sin‌230∘

2×10

=‌

1600×1

2×10×4

=20m
Also, at this time the horizontall distance travelled will be half of the range =OA i.e ‌

R

2

OA‌=‌

R

2

=‌

u2sin‌2θ

2g

=‌

402sin‌(2×30∘)

2×10

‌=‌

1600sin‌60∘

20

∴‌‌‌

R

2

=80‌

√3

2

=40√3 .
∴ The displacement from the projection point,
‌OB=D=√OA2+AB2
(By Pythagorous theorem)
⇒‌‌D=√(40√3)2+(20)2=√4800+400=√5200
⇒‌‌D=√4×13×100=2×10√13=20√13m