The cutaway of the gun barrel shows a projectile that upon exit moves with a spe

Question

The cutaway of the gun barrel shows a projectile that upon exit moves with a speed v=5490ft/s relative to the gun barrel. The length of the gun barrel is L=15ft. Assuming that the angle theta is increasing at a constant rate of 0.15 rad/s, determine the speed of the projectile right when it leaves the barrel. In addition, assuming that the projectile acceleration along the barrel is constant and that the projectile starts from rest, determine the magnitude of the acceleration upon exit. Please show ALL steps and will rate high for good answer and showing of steps.

Explanation / Answer

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there will be a tangent component of the speed of the projectile which is given by the formula w*r=2.25 ft/s
speed in the normal direction=5490 ft/s
so the magnitude of the total speed =(2.25^2+5490^2)^0.5=5490.000461 ft/s

for the acceleration
initial velocity=0
final velocity=5490.000461 ft/s
and length=15 ft
so v^2=u^2+2as
so a=1004670 m/s^2 is the required acceleration.

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