Question

a bullet is fired horizontally in the north direction with a velocity of 500m/sec at 30 N latitude calculate the horizontal component of the coriolis acceleration anf the consequent deflection of the bullet as it hits a target 250 m away .also find the vertically displacement of the bullet due to gravity​

Answer 1

$\Large\bf{\color{indigo}GiVeN,}$

• A bullet is fired horizontally in the north direction.

• Velocity of bullet = 500 m/s

• Bullet is inclined at 30° to the horizontal.

$\Large\bf\pink{To\:FiNd,}$

• The horizontal component of coriolis acceleration.

$\Large\bf\orange{CaLcUlAtIOn,}$

▪See the attachment diagram.

✒ If x-axis is taken vertically, z-axis towards north and y-axis along east.

$\Large\bf\red{We\:know\:that,}$

$\orange\bigstar\:\:\bf\purple{Angular\:Velocity \:=\:\omega\:\Big(\:\hat{k}\:\cos{30°}\:+\:\hat{\imath}\:\sin{30°}\:\Big)\:}$

☆ Because the angular velocity of the earth is directed parallel to it's axis and is inclined at 30° to the horizontal.

$\Large\bf\blue{We\:have,}$

$\pink\longmapsto\:\:\bf\green{\omega\:=\:\dfrac{2\pi}{Time\:period}\:}$

$\longmapsto\:\:\bf{\omega\:=\:\dfrac{2\pi}{24\times{60}\times{60}}\:}$

$\longmapsto\:\:\bf{\omega\:=\:7.2\times{10^{-5}}\:rad/sec\:}$

$\Large\bf\red{Hence,}$

$\green\longmapsto\:\:\bf\blue{Coriolis\:acceleration \:=\:-\:2\omega\times{v}\:}$

$:\implies\:\:\bf{Coriolis\:acceleration \:=\:-\:2\omega\:\Big(\:\hat{k}\:\cos{30°}\:+\:\hat{\imath}\:\sin{30°}\:\Big)\:\times{500\hat{k}}\:}$

$:\implies\:\:\bf{Coriolis\:acceleration \:=\:-\:2\times{7.2\times{10^{-5}}}\:\Big(\:\hat{k}\:\cos{30°}\:\times{500\hat{k}}\:+\:\hat{\imath}\:\sin{30°}\times{500\hat{k}}\:\Big)\:}$

$:\implies\:\:\bf{Coriolis\:acceleration \:=\:-\:2\times{7.2\times{10^{-5}}}\:\Big(\:0\:+\:\dfrac{1}{2}\times{500}\:\hat{\jmath}\:\Big)\:}$

$:\implies\:\:\bf{Coriolis\:acceleration \:=\:-\:2\times{7.2\times{10^{-5}}}\:\times{250}\:\hat{\jmath}\:}$

$:\implies\:\:\bf{Coriolis\:acceleration \:=\:0.036\:(-\hat{\jmath})\:m/s^2}$

$:\implies\:\:\bf\purple{Coriolis\:acceleration \:=\:(0.036\:m/s^2)\:towards\:west}$

$\Large\bf\therefore$ The horizontal component of coriolis acceleration is "0.036 $\bold\bf{m/s^2}$ towards west".