From the top of a tower of height $78.4m$ two stones are projected horizontally with $10m/s$ and $20m/s$ in opposite directions. On reaching the ground, their separation is

The correct option is B $120m$
For $I^{st}$ stone $u_h=10m/s$
$u_v=0;s_v=-78.4m;g=-9.8m/s^2$
$s_v=u_{v}t+\frac{1}{2}a{t}^2$

$\Rightarrow -78.4=\frac{-9.8}{2}\times t^2$
$t^2=16sec.\Rightarrow t=4sec.$

Distance travelled in horizontal direction in $4$ s

$s_n=10\times 4=40m$

For $I{I}^{nd}$ stone $\to u_h=20m/s;u_v=0;s_v=-78.4m;g=-9.8m/s^2$

It will also take same time as the vertical distance travelled is same. Distance travelled in horizontal direction by $I{I}^{nd}$ stone in $4$ s

$s_n=20\times 48=80m$

total distance between stones =$40+80=120m$