To solve this problem, we need to relate the maximum horizontal distance the ball can be thrown to the maximum height it reaches when thrown vertically.

Given Information:

• Maximum horizontal distance ($R$) = 52 m
• Initial speed ($v_0$) is the same for both horizontal and vertical throws.

Horizontal Throw Analysis:

The maximum range $R$ for projectile motion is given by the formula:

$R = \frac{v_0^2 \sin 2\theta}{g}$

where:

• $v_0$ is the initial speed,
• $\theta$ is the angle of projection,
• $g$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$).

For maximum range, $\theta = 45^\circ$. So, $\sin 2\theta = \sin 90^\circ = 1$.

Thus, the equation simplifies to:

$R = \frac{v_0^2}{g}$

Plugging in the values, we get:

$52 = \frac{v_0^2}{9.8}$

Solving for $v_0$:

$v_0^2 = 52 \times 9.8 = 509.6 \, \text{m}^2/\text{s}^2$ $v_0 = \sqrt{509.6} \approx 22.57 \, \text{m/s}$

Vertical Throw Analysis:

When the ball is thrown straight up, the maximum height ($H$) it reaches is given by:

$H = \frac{v_0^2}{2g}$

Substitute the value of $v_0^2 = 509.6$:

$H = \frac{509.6}{2 \times 9.8}$ $H = \frac{509.6}{19.6} \approx 26 \, \text{m}$

Thus, the maximum height the boy can throw the ball vertically upward is approximately 26 meters.

Would you like more details or have any questions? Here are some related questions you might find interesting:

1. How would the maximum height change if the acceleration due to gravity were different?
2. What would the maximum range be on the Moon, where gravity is 1/6th of Earth's gravity?
3. How long does the ball stay in the air when thrown vertically?
4. How much time does the ball take to reach its maximum height when thrown vertically?
5. What is the speed of the ball at its highest point when thrown vertically?

Tip: The angle of $45^\circ$ is ideal for achieving the maximum range in projectile motion, assuming no air resistance.