A bomb is dropped on an enemy post by an aeroplane flying horizontally with a velocity of $$60\,km\,{h^{ - 1}}$$ and at a height of $$490\,m.$$ At the time of dropping the bomb, how far the aeroplane should be from the enemy post so that the bomb may directly hit the target ?
A.
$$\frac{{400}}{3}m$$
B.
$$\frac{{500}}{3}m$$
C.
$$\frac{{1700}}{3}m$$
D.
$$498\,m.$$
Answer :
$$\frac{{500}}{3}m$$
Solution :
Time taken for vertical direction motion
$$t = \sqrt {\frac{{2h}}{g}} = \sqrt {\frac{{2 \times 490}}{{9.8}}} = \sqrt {100} = 10\,s$$
The same time is for horizontal direction.
$$\therefore x = vt = \left( {60 \times \frac{5}{{18}}} \right) \times 10 = \frac{{500}}{3}m$$
Releted MCQ Question on Basic Physics >> Kinematics
Releted Question 1
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