Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

An aeroplane is flying horizontally with a velocity $u = 600\text{ km/h}$ at a height of $1960\text{ m}$ . When it is vertically at a point $A$ on the ground, a bomb is released from it. The bomb strikes the ground at point $B$ . The distance $AB$ is:

1200 m

0.33 km

3.33 km

33 km

Step-by-Step Solution

Analyze Motion: The bomb dropped from the aeroplane undergoes projectile motion. It retains the horizontal velocity of the plane but has an initial vertical velocity of zero relative to the ground .
Convert Units: Horizontal velocity ( $u_x$ ) = $600\text{ km/h} = 600 \times \frac{5}{18}\text{ m/s} = \frac{500}{3}\text{ m/s}$ . Height ( $h$ ) = $1960\text{ m}$ .

Acceleration due to gravity ( $g$ ) is taken as $9.8\text{ m/s}^2$ (since 1960 is a multiple of 9.8).

Calculate Time of Flight ( $t$ ): Using the vertical motion equation $h = u_y t + \frac{1}{2}gt^2$ with $u_y = 0$ : $1960 = 0 + \frac{1}{2}(9.8)t^2$ $t^2 = \frac{1960 \times 2}{9.8} = \frac{3920}{9.8} = 400$ $t = \sqrt{400} = 20\text{ s}$ .
Calculate Horizontal Distance ( $AB$ ): The horizontal distance (Range) is covered at a constant velocity: $R = u_x \times t = \frac{500}{3} \times 20 = \frac{10000}{3}\text{ m} \approx 3333.33\text{ m}$ .
Final Conversion: $3333.33\text{ m} \approx 3.33\text{ km}$ .

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