JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Heights and Distances

The angle of elevation of a jet plane from a point A on the ground is

6 0^{\circ}

. After a flight of 20 seconds at the speed of 432 km / hour, the angle of elevation changes to

3 0^{\circ}

. If the jet plane is flying at a constant height, then its height is:

Answer: (C)

Solution

Speed = 432 km/h =

432 \times \frac{5}{18}

m/s = 120 m/s.

Distance traveled in 20 seconds =

120 \times 20 = 2400

m.

Let the height be

h

. Initially, from point A,

tan 6 0^{\circ} = \frac{h}{x _{1}}

, so

x_{1} = \frac{h}{3}

.

After flying 2400 m horizontally,

tan 3 0^{\circ} = \frac{h}{x _{2}}

, so

x_{2} = h 3

.

The horizontal distance covered is

x_{2} - x_{1} = 2400

h 3 - \frac{h}{3} = 2400

h (3 - \frac{1}{3}) = 2400

h \cdot \frac{3 - 1}{3} = 2400

h \cdot \frac{2}{3} = 2400

h = 12003

JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Single correct option