At what angle with the horizontal should a player throw a ball so that it may go to (i) a maximum distance (ii) reach maximum height?

Answer

To solve the problem of determining the angle at which a player should throw a ball to achieve maximum distance and maximum height, we can break it down into two parts:

### Part (i): Maximum Distance

1. **Understanding Projectile Motion**: When a projectile is thrown, it follows a parabolic path. The horizontal and vertical components of the initial velocity can be defined as:
- \( u_x = V_0 \cos(\theta_0) \) (horizontal component)
- \( u_y = V_0 \sin(\theta_0) \) (vertical component)

2. **Formula for Range**: The range \( R \) of a projectile is given by the formula:
\[ R = \frac{V_0^2 \sin(2\theta_0)}{g} \]
where \( g \) is the acceleration due to gravity.

3. **Maximizing the Range**: To maximize the range, we need to maximize \( \sin(2\theta_0) \). The maximum value of \( \sin(2\theta_0) \) is 1, which occurs when:
\[ 2\theta_0 = 90^\circ \quad \Rightarrow \quad \theta_0 = 45^\circ \]

4. **Conclusion for Maximum Distance**: Therefore, to achieve maximum distance, the angle should be:
\[ \theta_0 = 45^\circ \]

### Part (ii): Maximum Height

1. **Understanding Maximum Height**: The maximum height \( H \) reached by a projectile is given by the formula:
\[ H = \frac{u_y^2}{2g} = \frac{(V_0 \sin(\theta_0))^2}{2g} \]

2. **Maximizing the Height**: To maximize the height, we need to maximize \( \sin^2(\theta_0) \). The maximum value of \( \sin(\theta_0) \) is 1, which occurs when:
\[ \theta_0 = 90^\circ \]

3. **Conclusion for Maximum Height**: Therefore, to achieve maximum height, the angle should be:
\[ \theta_0 = 90^\circ \]

### Final Answers:
- For maximum distance: \( \theta_0 = 45^\circ \)
- For maximum height: \( \theta_0 = 90^\circ \)

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