Question

An aeroplane is flying in a horizontal direction with a velocity of $600km{h}^{-1}$ and at a height of $1960m$. When it is vertically above the point $A$, on the ground, a body is dropped from it. The body strikes the ground at point $B$. Then distance $AB$ is

• A. 5.8 km
• B. 4.7 km
• C. 3.3 km
• D. 2.0 km

Answer 1

Answer: (C)

The velocity of plane in horizontal direction,
$v_x=600km{h}^{-1}=\frac{600\times 1000}{60\times 60}m{s}^{-1}=\frac{500}{3}m{s}^{-1}$
Due to inertia this is also the velocity of body which remains constant during the flight of body.
Initial velocity of body in vertical direction $U_y=0$
If $t$ is time taken by the body to reach the earth, then from relation
$s=ut+\frac{1}{2}a{t}^2$, we have
$h=\frac{1}{2}g{t}^2\text{ or }t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times 1960}{9.8}}=20s$
$\therefore$ Distance traversed by body in horizontal direction
$AB=v_{x}t=\frac{500}{3}\times 20=\frac{10}{3}\times 10^{3}m$
$=3.3km$