A cricketer can throw a ball to the maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball?

1. Break Down the Problem

To determine the maximum height the cricketer can throw the ball, we can analyze the projectile motion of the ball. The problem presents a scenario involving horizontal distance and maximum height. We need to consider the basic principles of projectile motion.

2. Relevant Concepts

In projectile motion, the following relationships are important:

• The horizontal range $R$ is given by the formula: $R = \frac{v^2 \sin(2\theta)}{g}$ where $v$ is the initial velocity, $\theta$ is the angle of projection, and $g$ is acceleration due to gravity (approximately $9.81 \, \text{m/s}^2$).

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

3. Analysis and Detail

1. Since the maximum range is given as $R = 100 \, m$, we first determine the angle for maximum range, which is $\theta = 45^\circ$.

2. At this angle, the sine and cosine values are:

   • $\sin(45^\circ) = \cos(45^\circ) = \frac{\sqrt{2}}{2}$

3. Substitute $\theta = 45^\circ$ into the range formula to find $v$: $100 = \frac{v^2 \sin(90^\circ)}{9.81} \implies 100 = \frac{v^2}{9.81}$ $v^2 = 100 \times 9.81 = 981 \implies v = \sqrt{981} \approx 31.32 \, \text{m/s}$

4. Now we can find the maximum height $H$: $H = \frac{(31.32)^2 \sin^2(45^\circ)}{2 \times 9.81}$ $H = \frac{981 \left(\frac{1}{2}\right)}{19.62} \approx \frac{490.5}{19.62} \approx 24.99 \, \text{m}$

4. Verify and Summarize

The calculations were verified at each step, showing that the highest point the ball can reach is consistent through the application of projectile motion formulas.

Final Answer

The cricketer can throw the ball approximately 25.0 meters high above the ground.