A man can throw a stone 80 m. The maximum height to which he can raise the stone is

The correct option is 40 m.

To find the maximum height to which the man can raise the stone, we need to analyze the physics of projectile motion and vertical motion.

First, let's consider the maximum horizontal range. The horizontal range $R$ of a projectile launched with an initial velocity $u$ at an angle $\theta$ with the horizontal is given by the formula:
$R=\frac{u^{2}sin\left(2\theta \right)}{g}$
where $g$ is the acceleration due to gravity.

The range is maximum when $sin\left(2\theta \right)=1$, which occurs at a launch angle of $\theta ={45}^{\circ }$. The maximum horizontal range $R_{max}$ is:
$R_{max}=\frac{u^2}{g}$
Given that the man can throw the stone a maximum distance of 80 m, we have:
$R_{max}=\frac{u^2}{g}=80\text{m}$

Now, we want to find the maximum height $H$ to which the man can raise (throw vertically upwards) the stone. When throwing the stone straight up, the angle of projection is ${90}^{\circ }$. The maximum height $H$ reached by a stone thrown vertically upwards with the same initial velocity $u$ is given by the kinematic equation:
$v^2=u^2-2gH$
At the highest point, the final velocity $v=0$:
$0=u^2-2gH$
Solving for $H$:
$H=\frac{u^2}{2g}$

We can write this in terms of the maximum range $R_{max}$:
$H=\frac{1}{2}\left(\frac{u^2}{g}\right)=\frac{R_{max}}{2}$

Substituting the given value of $R_{max}=80\text{m}$ into the equation:
$H=\frac{80}{2}=40\text{m}$

Therefore, the maximum height to which the man can raise the stone is 40 m.