This is a projectile motion problem. Let's solve it step-by-step:

Problem Breakdown:

• Vertical height to reach: $h = 20.0 \, \text{m}$
• Launch angle: $\theta = 55.0^\circ$
• Acceleration due to gravity: $g = 9.8 \, \text{m/s}^2$
• Find initial velocity: $v_0$

Vertical Motion Equation:

The ball reaches the maximum height at the window. Using the vertical motion equation: $y = v_0 \sin\theta \cdot t - \frac{1}{2} g t^2$ At the maximum height ($y = 20.0$): $20.0 = v_0 \sin(55.0) \cdot t - \frac{1}{2} \cdot 9.8 \cdot t^2$

Time to Maximum Height:

From vertical velocity at the maximum height: $v_y = v_0 \sin\theta - g \cdot t$ At the maximum height ($v_y = 0$): $t = \frac{v_0 \sin\theta}{g}$

Substitute $t$ into the height equation: $20.0 = v_0 \sin(55.0) \cdot \frac{v_0 \sin(55.0)}{9.8} - \frac{1}{2} \cdot 9.8 \cdot \left(\frac{v_0 \sin(55.0)}{9.8}\right)^2$

Simplify the Expression:

$20.0 = \frac{(v_0^2 \sin^2(55.0))}{9.8} - \frac{4.9 \cdot (v_0^2 \sin^2(55.0))}{9.8^2}$ Factor out $v_0^2 \sin^2(55.0)$: $20.0 = \frac{v_0^2 \sin^2(55.0)}{9.8} \left(1 - \frac{4.9}{9.8}\right)$ $20.0 = \frac{v_0^2 \sin^2(55.0)}{9.8} \cdot \frac{1}{2}$ $v_0^2 = \frac{20.0 \cdot 9.8 \cdot 2}{\sin^2(55.0)}$ $v_0^2 = \frac{392.0}{\sin^2(55.0)}$

Solve for $v_0$:

Using $\sin(55.0) \approx 0.8192$: $v_0^2 = \frac{392.0}{0.8192^2}$ $v_0^2 = \frac{392.0}{0.6711}$ $v_0^2 \approx 584.0$ $v_0 \approx 24.2 \, \text{m/s}$

Final Answer:

The boy should throw the ball at an initial velocity of approximately 24.2 m/s.

Would you like a breakdown of any step, or more explanation? Here are some related questions:

1. How do you find the horizontal range of the ball?
2. What is the total time the ball is in the air?
3. How does changing the angle affect the required velocity?
4. What is the velocity of the ball as it reaches the window?
5. How does air resistance affect this calculation?

Tip: Always break motion problems into horizontal and vertical components to simplify the calculations.