An aeroplane is flying horizontally with a velocity of $600 km/hr$ and at a height of $2, 000 m$ . When it is vertically above a point $A$ on the ground, a bomb is released from it. The bomb strikes the ground at point $B$ . The distance $A B$ (in kilometer) is (neglecting air resistance, take
$g = 10 {m/s}^{2}$ )

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Solution

Taking downward direction as positive,
Using, $s_{y} = u_{y} + \frac{1}{2} a_{y} t^{2}$
$h = 0 + \frac{1}{2} g t^{2}$
$\Rightarrow t = \sqrt{\frac{2 h}{g}} = \sqrt{\frac{2 \times 2000}{10}} = 20 s$
Now,
$A B = s_{x} = u_{x} t + \frac{1}{2} a_{x} t^{2}$
$\Rightarrow s_{x} = (600 \times \frac{5}{18}) 20 + 0$
$(∵ a_{x} = 0)$
$\Rightarrow s_{x} = 3333.33 m$
$= 3.33 km$

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An aeroplane is flying horizontally with a velocity of 600 km/hr and at a height of 2,000 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. The distance AB (in kilometer) is (neglecting air resistance, take g=10 m/s2 )

Taking downward direction as positive, Using, sy=uy+12ayt2 h=0+12gt2 ⇒t=√2hg=√2×200010=20 s Now, AB=sx=uxt+12axt2 ⇒sx=(600×518)20+0 (∵ax=0) ⇒sx=3333.33 m =3.33 km