A cricketer can throw a ball to a maximum horizontal distance of 100 m. With the same effort, he throws the ball vertically upwards. The maximum height attained by the ball is

The correct option is 50 m.

Let us break down the physical concepts and mathematical steps to find the relationship between the maximum horizontal distance and the maximum vertical height.

Step 1: Understand the formula for maximum horizontal range
When a projectile is thrown with an initial velocity $u$ at an angle $\theta$ to the horizontal, its horizontal range $R$ is given by the formula:
$R=\frac{u^{2}sin\left(2\theta \right)}{g}$
where $g$ is the acceleration due to gravity.

The horizontal range is maximum when $sin\left(2\theta \right)=1$, which occurs when $\theta ={45}^{\circ }$.
Therefore, the maximum horizontal range $R_{\text{max}}$ is:
$R_{\text{max}}=\frac{u^2}{g}$

Given in the problem, the cricketer can throw the ball to a maximum horizontal distance of 100 m:
$R_{\text{max}}=\frac{u^2}{g}=100\text{m}$

Step 2: Understand the formula for maximum vertical height
With the same effort, meaning the same initial velocity $u$, the cricketer throws the ball vertically upwards.
For a vertical throw, the angle of projection with the horizontal is $\theta ={90}^{\circ }$.
The maximum height $H$ attained by a projectile is given by:
$H=\frac{u^2{sin}^2\left(\theta \right)}{2g}$

For a ball thrown vertically upwards, $sin\left({90}^{\circ }\right)=1$, so:
$H_{\text{max}}=\frac{u^2}{2g}$

Step 3: Relate the maximum height to the maximum range
By comparing the two expressions, we can write:
$H_{\text{max}}=\frac{1}{2}\left(\frac{u^2}{g}\right)=\frac{R_{\text{max}}}{2}$

Substitute the given value of $R_{\text{max}}=100\text{m}$:
$H_{\text{max}}=\frac{100}{2}=50\text{m}$

Thus, the maximum height attained by the ball is 50 m.