A ball is dropped from the top of a tower $120\text{m}$ high and simultaneously another ball is projected vertically upwards from the ground with a speed of $30\text{m/s}$. The two balls will meet after

The correct option is C $4\text{s}$
Let ${}^{\prime }{x}^{\prime }$ be the distance from the top of tower, where the balls projected will meet
$\therefore$ For ball freely falling body we know that the displacement covered is $s=\frac{1}{2}a{t}^2$, where $t$ is the time when both balls meet.

$\Rightarrow x=\frac{1}{2}g{t}^2...\left(i\right)$

For ball projected upwards: the displcement is $(120-x)$
$120-x=30t-\frac{1}{2}g{t}^2$
$\therefore$ From $\left(i\right)$ and $\left(ii\right)$

$\Rightarrow 120-\frac{1}{2}g{t}^2=30t-\frac{1}{2}g{t}^2$

$30t=120$
$t=4\text{s}$

Hence option C is the corect answer