Two paper screens A and B are separated by 15 | vedclass

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(D) The bullet exhibits horizontal projectile motion.Vertical displacement of the bullet,$y = 15 \ cm = 0.15 \ m$.Horizontal displacement of the bulle

Vedclass · Accepted Answer

$500 \sqrt{3} \ m \ s^{-1}$

Vedclass · Answer

(D) The bullet exhibits horizontal projectile motion.Vertical displacement of the bullet,$y = 15 \ cm = 0.15 \ m$.Horizontal displacement of the bullet,$x = 150 \ m$.Using the equation of trajectory for horizontal projection,$y = \frac{g x^2}{2 u^2}$.Substituting the values: $0.15 = \frac{10 	imes 150^2}{2 u^2}$.Rearranging for $u^2$: $u^2 = \frac{10 	imes 22500}{2 	imes 0.15} = \frac{225000}{0.3} = 750000$.Taking the square root: $u = \sqrt{750000} = \sqrt{250000 	imes 3} = 500 \sqrt{3} \ m \ s^{-1}$.

Two paper screens $A$ and $B$ are separated by $150 \ m$. $A$ bullet pierces $A$ and then $B$. The hole in $B$ is $15 \ cm$ below the hole in $A$. If the bullet is traveling horizontally at the time of hitting $A$,then the velocity of the bullet at $A$ is: $(g=10 \ m \ s^{-2})$

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