Published by:
CGP EDU Academic Team
Published on: September 12, 2026
An aeroplane is flying at a constant height of 1960 m with speed 
above the ground towards point directly over a person struggling in flood water. At what angle of sight with the vertical should be pilot release a survival kit if it is to reach the person in water? 
Text Solution
Verified by ExpertsThe correct answer is:
B
Step 1: Given the height of the airplane, $h = 1960 ext{ m}$ and its speed, $v = 600 ext{ km/h} = \frac{600}{3.6} = 166.67 ext{ m/s}$.
Step 2: The time $t$ it takes for the kit to fall can be calculated using the formula for free fall: $h = \frac{1}{2} g t^2$. Rearranging gives: $t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 1960}{9.8}} = \sqrt{400} = 20 ext{ s}$.
Step 3: During this time, the horizontal distance $x$ the airplane travels is: $x = v \times t = 166.67 \times 20 = 3333.33 ext{ m}$.
Step 4: To find the angle of sight, we use $ an(\phi) = \frac{x}{h} = \frac{3333.33}{1960}$. Therefore, $\phi = \tan^{-1}(1.70) \approx 59.04°$.
Thus, the angle of sight with the vertical is approximately 60°, giving option B.
Step 2: The time $t$ it takes for the kit to fall can be calculated using the formula for free fall: $h = \frac{1}{2} g t^2$. Rearranging gives: $t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 1960}{9.8}} = \sqrt{400} = 20 ext{ s}$.
Step 3: During this time, the horizontal distance $x$ the airplane travels is: $x = v \times t = 166.67 \times 20 = 3333.33 ext{ m}$.
Step 4: To find the angle of sight, we use $ an(\phi) = \frac{x}{h} = \frac{3333.33}{1960}$. Therefore, $\phi = \tan^{-1}(1.70) \approx 59.04°$.
Thus, the angle of sight with the vertical is approximately 60°, giving option B.
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