Question

Two paper screens $A$ and $B$ are separated by $150\text{m}$. A bullet pierces $A$ and $B$ such that hole in $B$ is $h=15\text{cm}$ below the hole in $A$. If the bullet is travelling horizontally at the time of hitting $A$, then the velocity of bullet at $A$ is

• A. $100\sqrt{3}m{s}^{-1}$
• B. $200\sqrt{3}m{s}^{-1}$
• C. $400\sqrt{3}m{s}^{-1}$
• D. $500\sqrt{3}m{s}^{-1}$

Answer 1

The correct option is D $500\sqrt{3}m{s}^{-1}$


Here bullet is travelling horizontally at $A$, therefore its vertical component of velocity $\left(u_y\right)$ is zero.
Let time taken by bullet to fall $h=15\text{cm}$ between A and B is $t$
Using equation of motion for vertical motion of bullet. (Taking downward motion $+ve$)

$h=u_{y}t+\frac{1}{2}a{t}^2\frac{15}{100}=0+\frac{1}{2}\times 10\times t^2$
$\frac{3}{100}=t\frac{\sqrt{3}}{10}=t$

With this time bullet reaches to B
So, $d=t\times u_{x}150=u_x\times \frac{\sqrt{3}}{10}{u}_x=\frac{1500}{\sqrt{3}}\text{m/s}=500\sqrt{3}\text{m/s}$