At $t=0$, an arrow is fired vertically upwards with a speed of $100{\text{ms}}^{-1}$. A second arrow is fired vertically upwards with the same speed at $t=5\text{s}$. Then (take $g=10m{s}^{-2}$)(Neglect air friction)

The correct options are
A the two arrows will be at the same height above the ground at $t=12.50\text{s}$
B the two arrows will reach back their starting points at $t=20\text{s}$ and $t=25\text{s}$ respectively
C the ratio of the speeds of the first and second arrows at $t=20\text{s}$ will be $2:1$
Let them meet at height $h$ after time $t$.
$h=100t-\frac{1}{2}g{t}^2\to$ for first arrow
$h=100(t-5)-\frac{1}{2}g(t-5{)}^2\to$ for second arrow
$\Rightarrow t=12.5$ (after solving). So (a) is correct.
Time of flight of first arrow: $T=\frac{2u}{g}=\frac{2\times 100}{10}=20\text{s}$
Second arrow will reach after $5\text{s}$ of reaching first, so (b) is correct
$v_1=100-10\times 20=-100\text{m/s}$
$v_2=100-10\times 15=-50\text{m/s}$
Ratio: $\frac{v_1}{v_2}=2:1$, so (c) is correct.
Maximum height attained
$H=\frac{u^2}{2g}=\frac{(100{)}^2}{2\times 10}=500\text{m}$.
Hence (d) is incorrect.