Question

An aeroplane is flying in a horizontal direction with a velocity of $ 600\text{ }km{{h}^{-1}} $ and at a height of $1960\, m$. 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,
$v_{x}=600 \,kmh ^{-1}=\frac{600 \times 1000}{60 \times 60} \,ms ^{-1}=\frac{500}{3}\, ms ^{-1}$
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
$s=u t+\frac{1}{2} a t^{2}$, we have
$h=\frac{1}{2} g t^{2} \text { or } t=\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2 \times 1960}{9.8}}=20 s$
$\therefore $ Distance traversed by body in horizontal direction
$A B=v_{x} t=\frac{500}{3} \times 20=\frac{10}{3} \times 10^{3} m$
$=3.3\, km$