Two bullets are fired horizontally and simultaneously towards each other from roof tops of two buildings $100 \mathrm{~m}$ apart and of same height of $200 \mathrm{~m}$ with the same velocity of $25 \mathrm{~m} / \mathrm{s}$. When and where will the two bullets collides. $\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right)$

Given, distance between the two buildings $\mathrm{d}=100 \mathrm{~m}$ height of each tower, $\mathrm{h}=200 \mathrm{~m}$ speed of each bullet, $\mathrm{v}=25 \mathrm{~ms}^{-1}$ The situation can be shown as below
where, $\mathrm{x}$ be the vertical distance travelled from the top of the building and t be the time at which they collide. As two bullets are fired toward each other, So, their relative velocity will be $$ \mathrm{v}_{\text {rel }}=25-(-25)=50 \mathrm{~ms}^{-1} $$ Then, time $\mathrm{t}=\frac{\mathrm{d}}{\mathrm{v}_{\text {rel }}}=\frac{100}{50}=2 \mathrm{~s}$ The distance or height at which they collide is calculated from equation of motion, $$ \mathrm{x}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^2 $$ The bullet is initially a rest i.e. $\mathrm{u}=0$ and $\mathrm{as}$ it is moving under the effect of gravity $a=-g$, so $$ \begin{aligned} & x=-\frac{1}{2} g^2 \\ & x=-\frac{1}{2} \times 10(2)^2=-20 \mathrm{~m} \end{aligned} $$ The negative sign shows that the bullets will collide $20 \mathrm{~m}$ below the top of tower i.e. at a height of $(200-20)=180 \mathrm{~m}$ from the ground after $2 \mathrm{~s}$.