If time of flight of a projectile is 10 seconds. Range is 500 m. The maximum height attained by it will be

The correct option is 125 m.

Let us find the maximum height attained by the projectile step-by-step.

1. Identify the given values:
- Time of flight ($T$) = $10\text{s}$
- Horizontal range ($R$) = $500\text{m}$
- Acceleration due to gravity ($g$) is approximately $10{\text{m/s}}^2$.

2. Understand the formula for Time of Flight:
The time of flight of a projectile is given by the formula:
$T=\frac{2u\sin \theta }{g}$
where $u$ is the initial velocity and $\theta$ is the angle of projection. The vertical component of the initial velocity is $u_y=u\sin \theta$. Therefore, we can rewrite the time of flight as:
$T=\frac{2u_y}{g}$

3. Calculate the vertical component of the initial velocity ($u_y$):
Substitute the given values into the time of flight formula:
$10=\frac{2u_y}{10}$
Multiplying both sides by 10 gives:
$100=2u_y$
Dividing by 2:
$u_y=50\text{m/s}$

4. Calculate the Maximum Height attained ($H$):
The maximum height attained by a projectile depends only on the vertical component of its initial velocity and gravity. The formula is:
$H=\frac{u^2{\sin }^2\theta }{2g}=\frac{u_y^2}{2g}$
Now, substitute the value of $u_y=50\text{m/s}$ and $g=10{\text{m/s}}^2$ into the maximum height equation:
$H=\frac{{50}^2}{2\times 10}$
Simplify the expression:
$H=\frac{2500}{20}$
$H=125\text{m}$

Thus, the maximum height attained by the projectile is 125 m.