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Standard XII

Physics

Applications of Horizontal and Vertical Components

An aeroplane ...

Question

An aeroplane is flying horizontally with a velocity of $600 km/h$ at a height of $1960 m$ . When it is vertically above a point $A$ on the ground, a bomb is released from it. The bomb strikes the ground at point $B$ . If $g = 9.8 {m/s}^{2}$ then the distance $A B$ is

$1200 m$

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$0.33 km$

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$3.33 km$

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$33 km$

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Solution

The correct option is C
$3.33 km$

The time taken by a bomb to reach the ground falling from a height $h = 1960 m$ is given by

$t = \sqrt{\frac{2 h}{g}} = {(\frac{2 \times 1960}{9.8})}^{1 / 2} = 20 s t = \frac{20}{60 \times 60} h = \frac{1}{180} h$

$∴$ Distance $A B$
$A B = v t = 600 (km/h) \times \frac{1}{180} (h) = 3.33 km$

Hence, the correct choice is $(c)$ .

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An aeroplane is flying horizontally with a velocity of 600 km/h at a height of 1960 m. When it is vertically above a point A on the ground, a bomb is released from it. The bomb strikes the ground at point B. If g=9.8 m/s2 then the distance AB is

The correct option is C 3.33 km The time taken by a bomb to reach the ground falling from a height h=1960 m is given by t=√2hg=(2×19609.8)1/2=20 st=2060×60 h=1180 h ∴ Distance AB AB=vt=600 (km/h)×1180 (h)=3.33 km Hence, the correct choice is (c).

An aeroplane is flying horizontally with a velocity of $600 km/h$ at a height of $1960 m$ . When it is vertically above a point $A$ on the ground, a bomb is released from it. The bomb strikes the ground at point $B$ . If $g = 9.8 {m/s}^{2}$ then the distance $A B$ is