A projectile is thrown vertically with an initial velocity of 20ms^-1. Which of the following would be the vertical distance traveled by the projectile if it starts to fall after 2 seconds?

Acceleration due to gravity, $\mathrm{g}=9.8 {\mathrm{ms}}^{-2}$
The initial velocity of the projectile, u = 20 ms^-1

time = 2 sec
We know that,

$\mathrm{S}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2$
Here $\mathrm{a}=-\mathrm{g}$            [Acceleration is opposite to gravity]
Here we can write the above equation as $\mathrm{S}=\mathrm{ut}-\frac{1}{2}{\mathrm{gt}}^2----\left(\mathrm{i}\right)$
Substituting all the values in the equation (i)
We get  $\mathrm{S}=20\times 2-\frac{1}{2}\times 9.8{\times \left(2\right)}^2$
$\Rightarrow \mathrm{S}=40-\frac{39.2}{2}$ 

$\Rightarrow \mathrm{S}=40 -19.6=20.4 \mathrm{m}$ 

Thus, the vertical distance traveled by the projectile in 2 seconds is 20.4 m.