Question

Two bullets are fired horizontally and simultaneously towards each other from the rooftops of two buildings located $200\text{m}$ apart, ot the same height of $300\text{m}$ with the same velocity of $50\text{m/s}$. Which of the following is/are correct? [Take $g=10{\text{m/s}}^2$]

• A. They collide after $2\text{s}$.
• B. They collide after $1\text{s}$.
• C. They collide at a height of $290\text{m}$.
• D. They collide at a height of $280\text{m}$.

Answer 1

Answer: (A), (D)

The correct options are
A They collide after $2\text{s}$.
D They collide at a height of $280\text{m}$.

Figure as per question -


As there is no force on the bullets in the horizontal direction, the centre of mass of the two bullet system will not change.
Hence, we can say that, collision takes place at the position of the centre of mass.
So, distance of COM from the top of building $1$ (along horizontal direction) is given by
$x=\frac{m_1{d}_1+m_2{d}_2}{m_1+m_2}$
$\Rightarrow x=\frac{m\times 0+m\times d}{m+m}$
[ $m$ is the mass of the bullet ]
$\Rightarrow x=\frac{d}{2}$ ............(1)
Now, time taken by the bullet to reach this position from the top of building $1$ is given by
$t=\frac{x}{v_1}$
$\Rightarrow t=\frac{d}{2v_1}$ [ from (1) ]
$\Rightarrow t=\frac{200}{2\times 50}=2\text{s}$ [ from figure ]
In $2\text{s}$, the bullet will also travel along the vertical.
For vertical motion, $s=ut+\frac{1}{2}a{t}^2$
$\Rightarrow y=0+\frac{1}{2}\times 10\times 2^2=20\text{m}$
$\therefore$ Height from ground when they collide,
$h=300-y$ [ from figure ]
$=300-20=280\text{m}$

Option (a) and (d) are correct.