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Question

(D) The bullet travels horizontally,so its initial vertical velocity $u_y = 0$. Given: Horizontal distance $S_x = 100\,m$,vertical displacement $h = 1

Vedclass · Accepted Answer

$700$

Vedclass · Answer

(D) The bullet travels horizontally,so its initial vertical velocity $u_y = 0$. Given: Horizontal distance $S_x = 100\,m$,vertical displacement $h = 10\,cm = 0.1\,m$,and acceleration due to gravity $g = 10\,m/s^2$. Using the equation of motion in the vertical direction: $h = u_y t + \frac{1}{2} g t^2$. Since $u_y = 0$,we have $h = \frac{1}{2} g t^2$. Substituting the values: $0.1 = \frac{1}{2} 	imes 10 	imes t^2$. $0.1 = 5 t^2 \implies t^2 = 0.02 \implies t = \sqrt{0.02} \approx 0.1414\,s$. In the horizontal direction,the velocity $v_x$ is constant: $v_x = \frac{S_x}{t}$. $v_x = \frac{100}{0.1414} \approx 707\,m/s$. Rounding to the nearest provided option,the velocity is $700\,m/s$.

Two paper screens $A$ and $B$ are separated by a distance of $100\,m$. $A$ bullet pierces $A$ and then $B$. The hole in $B$ is $10\,cm$ below the hole in $A$. If the bullet is travelling horizontally at the time of hitting $A$,then the velocity of the bullet at $A$ is $.......\,m/s$.

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