A smart anti-armor weapon is launched off a fighter aircraft flying at 500 knots

Question

A smart anti-armor weapon is launched off a fighter aircraft flying at 500 knots straight and level at an altitude of 2000 feet above ground level. The weapon is relaesed from the aircraft with an initial velocity of 15 ft/s downward. The weapon will fly intact to 700 feet above ground level where it will open and begin to function. Neglecting air friction:

I am struggling to determine the time for the weapon to reach 700 ft above ground level (AGL); the total velocity at 700 ft AGL;

Any assistance is appreciated.

Explanation / Answer

using the equations of dynamics we have S = u*t + 1/2*a*t^2 and also V^2 - u^2 = 2*a*S where S =distance traveled t = time taken , a = acceleration and u = initial velocity. v = final velocity. here given u =15 ft/s a = acceleration due to gravity = 32.2 ft/s^2 and S = 700 ft above ground level = 2000-700 from the launching point = 1300 ft substituting in V^2- u^2 =2*a*S we have V = 289.73 ft/s now using V = u +a*t we have t = 8.532 s. now the total velocity of the weapon is sqrt(V^2 + x^2) where x= velocity of the aircraft launching the weapon. x = 500*1.68780 = 843.9 ft/s => the total velocity is V' = 892.25 ft/s

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