An aeroplane is flying horizontally with a velocity $u = 600 ext{ km/h}$ at a height of $1960 ext{ m}$. When

Question

An aeroplane is flying horizontally with a velocity $u = 600	ext{ km/h}$ at a height of $1960	ext{ m}$. 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:

Accepted Answer

1.  **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 .
2.  **Convert Units:**
    *   Horizontal velocity ($u_x$) = $600	ext{ km/h} = 600 	imes \frac{5}{18}	ext{ m/s} = \frac{500}{3}	ext{ m/s}$.
    *   Height ($h$) = $1960	ext{ m}$.
    *   Acceleration due to gravity ($g$) is taken as $9.8	ext{ m/s}^2$ (since 1960 is a multiple of 9.8).
3.  **Calculate Time of Flight ($t$):**
    Using the vertical motion equation $h = u_y t + \frac{1}{2}gt^2$ with $u_y = 0$:
    $1960 = 0 + \frac{1}{2}(9.8)t^2$
    $t^2 = \frac{1960 	imes 2}{9.8} = \frac{3920}{9.8} = 400$
    $t = \sqrt{400} = 20	ext{ s}$ .
4.  **Calculate Horizontal Distance ($AB$):**
    The horizontal distance (Range) is covered at a constant velocity:
    $R = u_x 	imes t = \frac{500}{3} 	imes 20 = \frac{10000}{3}	ext{ m} \approx 3333.33	ext{ m}$.
5.  **Final Conversion:**
    $3333.33	ext{ m} \approx 3.33	ext{ km}$.

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