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A rifle was aimed at the vertical line on the target located precisely in the northern direction, and then fired. Assuming the air drag to be negligible, find how much off the line, and in what direction, will the bullet hit the target. The shot was fired in the horizontal direction at the latitude varphi=60^@, the bullet velocity v=900m//s, and the distance from the target equals s=1.0km. |
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Answer» Solution :Define the axes as shown with z along the local vertical, x due EAST and y due north. (We assume we are in the northern hemisphere). Then the Coriolis FORCE has the componets. `vecF_(cor)=-2m(vecomegaxxvecv)` `=2momega[v_ycos theta-v_xsin theta)veci-v_xcos thetavecj+v_xcos thetaveck]=2momega(v_ycos theta-v_xsin theta)veci` since `v_x` is small when the direction in which the gun is fired is due north. Thus the equation of motion (neglecting centrifugal forces) are `overset(..)x=2momega(v_ysin varphi-v_xcos varphi)`, `overset(..)y=0` and `overset(..)z=-g` Integrating we GET `overset(.)y=V` (constant), `overset(.)z=-g t` and `overset(.)x=2omegavsin varphit+omegag t^2cos varphi` FINALLY, `x=omegavt^2sin varphi+1/3omega g t^3cos varphi` Now `vgt gtg t` in the present case. so, `x~~omegavsin varphi(s/v)^2=omega sin varphi(s^2)/(v)` `~~7cm` (to the east). 
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