A stone is dropped from the top of a $20m$ high tower. $1$sec later another stone is thrown downwards from the top. Both stone reach the ground simultaneously. If g is $10m/s^2$, initial velocity of second stone is

The correct option is B $15m/s$
$I$ dropped $\Rightarrow$ initial velocity $=0$

$\Rightarrow$ time will be given by $h=\frac{1}{2}g{t}^2$

$t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times 20m}{10m/s^2}}=2sec$

$II$ both reaches simultaneous be so ti,e taken by $IInd$ stone will be

$2sec-1sec=1sec$

or $t^{\prime }=1sec$

putting it in $h=ut$'$+\frac{1}{2}g{t}^2$

& $h=20m,g=10m/s^2$

we get

$20=u\times 1+5\times 1$

$\Rightarrow u=15m/s$