A rifle with a muzzle velocity of 2, 500 (ft/s) is fired at a slight up angle of

Question

A rifle with a muzzle velocity of 2, 500 (ft/s) is fired at a slight up angle of 1.25 degree. Once it leaves the barrel, the drag force in x causes a velocity dependent deceleration of: a_x = -0.0000120 middot v_x^2 (ft/sec^2) Since the bullet's vertical velocity is small throughout the motion its vertical acceleration is just the constant gravitational acceleration of: a_y = -32.2 (ft/sec^2) If the rifle is fired from an elevation of 4.38 (ft), and the bullet travels over level ground. How far will it travel before hitting the ground?

Explanation / Answer

step;1

Given that

velocity u=2500 ft/s

angle=1.25 degree

ay=-32.2 ft/s^2

step;2

now we accelration in x-axis

ax=(-0.000012)*(2500*sin1.25)^2=-0.04 ft/s^2

acceleration a=[0.04^2+32.2^2]^1/2=32.2 ft/s

now we find the range of the bullet travel

range R=(2500)*[2*4.38/32.2]^1/2=1303.9 ft

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