A cricketer can throw a ball to a maximum horizontal distance of 100 m. With the same speed, what is the maximum height the cricketer can throw the ball?

Step 1: Relate the maximum horizontal range to the initial speed. The maximum horizontal range ($R_{max}$) for a projectile launched with an initial speed $u$ occurs at a launch angle of $45^\circ$. The formula for maximum horizontal range is: $R_{max} = \frac{u^2}{g}$ Given $R_{max} = 100 \text{ m}$. $100 m = \frac{u^2}{g}$ From this, we can express $u^2$: $u^2 = 100g$

Step 2: Calculate the maximum vertical height. To achieve the maximum vertical height ($H_{max}$), the ball must be thrown straight upwards, meaning the launch angle is $90^\circ$. The formula for maximum vertical height is: $H_{max} = \frac{u^2}{2g}$ Substitute the expression for $u^2$ from Step 1 into this formula: $H_{max} = \frac{100g}{2g}$ $H_{max} = \frac{100}{2}$ $H_{max} = 50 m$

The maximum height the player can throw the ball is 50 m.

The final answer is $\boxed{50 m}$.