At time $t=0$, a bullet is fired vertically upwards with a speed of 98 $m{s}^{-1}.$ At time $t=5s$ (i.e., 5 seconds later) a second bullet is fired vertically upwards with the same speed. If the air resistance is neglected, which of the following statements is/are true?

The correct options are
A The two bullets will be at the same height above the ground at $t=12.5s$.
D The two bullets will attain the same maximum height.
For first bullet,
Time of ascent $=\frac{u}{g}=\frac{98}{9.8}=10s$ (Using first equation of motion)
$\therefore$ After $12.5s$, position of 1st bullet from the ground
$=98\times \frac{125}{10}-\frac{1}{2}\times 9.8\times (12.5{)}^2=459.375m$
As the second bullet is fired $5s$ later than 1st bullet, $t=(12.5-5)=7.5s$
Using second equation of motion, we get
$=98\times \frac{75}{10}-\frac{1}{2}\times \frac{98}{10}\times \frac{(75{)}^2}{100}=459.375m$

Hence, it is clear that position of both the bullets will be at same height at $t=12.5s$.

Maximum height in case of motion under gravity neglecting air resistance depends only on initial velocity. So the two bullets will attain the same maximum height.