A man is standing on top of a building 100 m high. He throws two balls vertically upward , one at $t=0$ and after a time interval (less than 2 s). The later ball is thrown at a velocity of half the first. The vertical gap between first and second ball is +15 m at $t=2$s. The gap is found to remain constant. Select the correct alternatives.

Let the speeds of the two balls (1 and 2) be $v_1$ and $v_2$, where if $v_1=2v, v_2=v$

If $y_1$ and $y_2$ are the distances covered by the balls 1 and 2 respectively. 

Then at the highest position,                          $y_1=\frac{v_1^2}{2g}=\frac{4v^2}{2g}$ and $y_2=\frac{v_2^2}{2g}=\frac{v^2}{2g}$

Since,                $y_1-y_2=15 \mathrm{m}, \frac{4v^2}{2g}-\frac{v^2}{2g}=15 \mathrm{m}$

or                              $\frac{3v^2}{2g}=15 \mathrm{m}$

or                                   $v^2=\sqrt{5\mathrm{m}\times (2\times 10)}$ m/s²

or                                     $v=10$ m/s

Clearly,                            $v_1=20$m/s and $v_2=10$ m/s

as                                    $y_1=\frac{v_1^2}{2g}=\frac{{\left(20\mathrm{m}\right)}^2}{2\times 10\mathrm{m}15}=20 \mathrm{m},$

                                         $y_2=y_1-15\mathrm{m}=5\mathrm{m}$

If $t_2$ is the time taken by the ball 2 to cover a distance of 5m, then from 0 = $v_2-g{t}_2, we get t_2=\frac{v_2}{g}=\frac{10}{10} s=1 s$

Since, $t_1$ (time taken by ball 1 to cover distance of 20m) =$\frac{v_2}{g}=\frac{20}{10} s=$ 2s, time interval between two throws

                                             $=t_1-t_2=2s-1s=1s$