A cricketer can throw a ball to a maximum horizontal distance of d. How high above the ground can the cricketer throw the same ball?d/2d2d5d/2

To find the maximum height a cricketer can throw a ball, we can use the concepts of projectile motion.

1. Break Down the Problem

We know that the maximum horizontal distance $d$ is achieved at an angle of $45^\circ$. We need to find the maximum height reached by the ball when thrown.

2. Relevant Concepts

• The horizontal distance (range) formula for projectile motion is given by: $R = \frac{v^2 \sin(2\theta)}{g}$ where $R$ is the range, $v$ is the initial velocity, $\theta$ is the angle of projection, and $g$ is the acceleration due to gravity.

• The maximum height $H$ reached by a projectile is given by: $H = \frac{v^2 \sin^2(\theta)}{2g}$

3. Analysis and Detail

1. Since $R = d$ and occurs at $\theta = 45^\circ$: $R = \frac{v^2}{g}$

2. From this, we can express $v^2$ as: $v^2 = d \cdot g$

3. Now, substituting $v^2$ back into the height formula at the angle of $90^\circ$ for maximum height: $H = \frac{d \cdot g \cdot \sin^2(90^\circ)}{2g} = \frac{d}{2}$

4. Verify and Summarize

Therefore, the maximum height that the ball can be thrown is $\frac{d}{2}$.

Final Answer

The maximum height above the ground that the cricketer can throw the ball is $\frac{d}{2}$.