During a target practice a warplane is moving at 100 m/s eastbound at an altitud

Question

During a target practice a warplane is moving at 100 m/s eastbound at an altitude of 1 km. It then accelerates at a rate of 20m/s^2, and after 10 seconds it fires a projectile horizontally with a speed of -50m/s respect to the airplane (assume the release is very quick). a) Does the projectile move forward or backwards according to: The airplane pilot? An observer on the ground? b) At what velocity does the projectile leave the airplane according to a ground observer? c) At what horizontal distance will the projectile hit? d) What will the velocity be at impact?

Explanation / Answer

Speed after 10 sec, v = v0 + a t = 100 + (20 x 10)

v = 300 m/s

(A) 1. backwards

2. forward

(b) v0 = 300 - 50 = 250 m/s

(C) it will leave horizontally,

so v0y = 0

y = 0 - 1000 = - 1000 m

ay = -9.8 m/s^2

Applying y= v0y t + ay t^2 /2

- 1000 = 0 - 9.8 t^2 /2

t = 14.3 sec

horizontal distance, D = 250 x 14.3

D = 3570 m Or 3.57 km

(d) vfy = v0y + ay t

vfy = 0 - (9.8 x 14.3) = 140 ms

vfx = 250 m/s

Speed = sqrt[ vfx^2 + vfy^2]

v = 287 m/s

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