Question

Two paper screens A and B are separated by a distance of $100m$. A bullet pierces A and then B. The hole in B is $10cm$ below the hole in A. If the bullet is travelling horizontally at A, velocity of the bullet at A is $(g=9.8m/s^2)$

• A. $500m/s$
• B. $700m/s$
• C. $800m/s$
• D. $900m/s$

Answer 1

Answer: (B)

$700m/s$

For vertical motion
$s=0.1m$, $g=9.8m/s^2$ and $u_v=0$

$s=u_{v}t+\frac{1}{2}g{t}^2\Rightarrow 0.1=(0\times t)+(\frac{1}{2}\times 9.8\times t^2)$
$\Rightarrow 0.1=4.9t^2\Rightarrow t=\frac{1}{7}sec$

For horizontal motion
$u_h=\frac{s}{t}$ (as acceleration is 0 in horizontal direction)

$u_h=\frac{100}{\left(\frac{1}{7}\right)}$

$\Rightarrow u_h=700m/s$