UPSC NDA Math Model Paper 5

Height of plane (vertically) = 5000 m
As shown in the figure, plane moves from 1 to 2 and man standing at M ∠O1M = 30° and ∠O2M = 60°
As per the △ O2M,

OM

O2

=tan‌60=√3
O2=

OM

√3

=

5000

√3

m
Similarly in Δ O2M,

OM

O1

=tan‌30=

1

√3

O1=5000√3m
Now as the plane moves from 1 to 2 in 30 sec.
Distance moved by the plane (d) = O1 - O2
⇒d=5000√3−

5000

√3

⇒d=

10000

√3

≈5773.5m
∴ Speed of plane (v)=

d

time

⇒v=

5773.5

30

≈192.45m∕s