At $t=0$ , an arrow is fired vertically upwards with a speed of $100m/s$ . A second arrow is fired vertically upwards with the same speed at $t=5s$ . Then $\left(g=10m/s^2\right)$

Let they meet at height $h$after time $t$ 

$h=100t-\frac{1}{2}g{t}^2\to$for first arrow

$=100\left(t-5\right)-\frac{1}{2}g{\left(t-5\right)}^2\to$for 2^nd arrow.

$\Rightarrow t=12.5s$ (after solving) so (1) is correct

$\Rightarrow$ Time of flight of first arrow

$T=\frac{2u}{g}=\frac{2\times 100}{10}=20s$

Second arrow will reach after $5s$ of reaching first

$T=25s$ $\left(2nd\right)$  correct

$\Rightarrow V_1=100-10\times 20=-100m/s$

$V_2=100-10\times 15=-50m/s$

$\frac{V_1}{V_2}=2:1$  (3) correct

$\Rightarrow H=\frac{u^2}{2g}=\frac{{\left(100\right)}^2}{2\times 10}=500m$   (4) incorrect