Two planes are flying at the same speed of 200m/sec in opposite directions. A machine-gun mounted in one plane fires at the other at right angle to their line of flight. How far apart will the bullet-holes made in the side of the second plane be, if the machine-gun fires 900 rounds per minute / What role does air-resistance play in this ?
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Sol. The aeroplanes are moving relative to each other at a speed equal to the sum of their speeds, i.e., at a speed, v of 400 m/sec. Between the firing of any two rounds a period of time, t, elapses =
min =
sec. The distance between the bullet-holes must equal the relative distance traveled by the second aeroplane during this time, i.e., S = vt =
= 27 m approx.
Since the length of the fuselage of an aeroplane rarely exceeds 27 m, not more than one bullet can normally hit the aeroplane the given conditions of fring.
As a result of air-resistance every bullet will require a greater length of time to traverse the distance between the two aeroplanes. But every bullet will be delayed by the same amount. Therefore the interval of time between the arrival at the target of any two consecutive bullets remains
sec as before, and the distance, between the bullet-holes must, as before, equal 27 m.
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