A particle is thrown vertically upwards. If its velocity at half of the maximum height is 10 m/s, then maximum height attained by it is (Take g =10 m/s²)

The correct option is 10 m.

Let the maximum height attained by the particle be $H$.

When the particle reaches its maximum height, its final velocity becomes zero. Let us analyze the motion of the particle from the point where it is at half of the maximum height to the top of its path.

According to the third equation of motion:
$v^2=u^2-2gs$
where:
- $u$ is the velocity of the particle at half of the maximum height, which is $10\text{m/s}$.
- $v$ is the velocity at the maximum height, which is $0\text{m/s}$.
- $g$ is the acceleration due to gravity, given as $10{\text{m/s}}^2$.
- $s$ is the vertical displacement from half of the maximum height to the maximum height, which is:
$s=H-\frac{H}{2}=\frac{H}{2}$

Substituting these values into the equation of motion:
$0^2={10}^2-2g\left(\frac{H}{2}\right)$
$0=100-gH$
$gH=100$

Substitute the value of $g=10{\text{m/s}}^2$ into the equation:
$10H=100$
$H=10\text{m}$

Therefore, the maximum height attained by the particle is 10 m.