A jet fighter travelling at 600 mph fires an 85 kg sidewinder missile. The missi

Question

A jet fighter travelling at 600 mph fires an 85 kg sidewinder missile. The missile falls for 3 seconds before its rocket engine ignites. The engine will produce 17 kN of thrust. A bit unrealistically, you may neglect air resistance.

(A) At the moment just after the missile engine is ignited what is the velocity of the rocket (magnitude and direction)

(B) At the same time what is the net force on the missile (magnitude and direction)?

(C) What is the acceleration then of the missile (magnitude and direction)?

(D) Modern air-to-air missiles, including the latest sidewinder, employ thrust vectoring, which allows the missile to "steer" before it gets going fast enough for the tiny tail fins to become effective.   If this missile were employing such thrust-vectoring technology, in what direction should the thrust push the missile if the target is at the same altitutde as the missile when the engine ignites (although far away)?   

Explanation / Answer

Here ,
a)just after the engine is fired ,

Vh = 600 mph = 268.2 m/s

vv = a * t

Vv = 9.8 * 3 = 29.2 m/s

Now , magnitude of velocity = sqrt(268.2^2 + 29.2^2)

magnitude of velocity = 269.8 m/s

direction = arctan(29.2/268.2)

direction = 6.21 degree below horizontal

B)

net force = 85 * 9.8 j + 17000 j N

net force = sqrt(833^2 + 17000^2) at arctan(833/17000) degree below horizontal

net force = 17020 N at 2.81 degree below horizontal

C)

using second law of motion

acceleration = net force/m

acceleration = (17020/81) at 2.81 degree below horizontal

acceleration = 210.12 m/s^2 at 2.81 degree below horizontal

D)

for the missile ,

as the net force must be horizontal

17000 * sin(theta) = 85 * 9.8

theta = 2.81 degree

the direction of the thrust is 2.81 degree above horizontal

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