To find the height of the building, we need to analyze the motion of the soccer ball. This is a projectile motion problem, where the horizontal distance (range), initial velocity, and angle of projection are given.

Step 1: Break down the initial velocity into components.

• The initial velocity $v_0 = 19 \, \text{m/s}$
• Angle of projection $\theta = 49^\circ$

The horizontal and vertical components of the velocity are: $v_{0x} = v_0 \cos(\theta) = 19 \cos(49^\circ)$ $v_{0y} = v_0 \sin(\theta) = 19 \sin(49^\circ)$

Step 2: Calculate the time of flight.

The horizontal distance (range) is given by: $\text{Range} = v_{0x} \times t$ where $t$ is the time of flight. Solving for $t$: $t = \frac{\text{Range}}{v_{0x}} = \frac{65}{19 \cos(49^\circ)}$

Step 3: Use the time of flight to find the height of the building.

The vertical displacement (height of the building) is given by: $\Delta y = v_{0y} \times t - \frac{1}{2} g t^2$ where $g = 9.8 \, \text{m/s}^2$ is the acceleration due to gravity.

Now, let's calculate this step by step.

1. Calculate $v_{0x}$ and $v_{0y}$: $v_{0x} = 19 \cos(49^\circ) \approx 12.47 \, \text{m/s}$ $v_{0y} = 19 \sin(49^\circ) \approx 14.34 \, \text{m/s}$

2. Calculate the time of flight $t$: $t = \frac{65}{12.47} \approx 5.21 \, \text{seconds}$

3. Calculate the height of the building: $\Delta y = 14.34 \times 5.21 - \frac{1}{2} \times 9.8 \times (5.21)^2$ $\Delta y \approx 74.76 - 133.13 \approx -58.37 \, \text{meters}$

Since the displacement $\Delta y$ is negative, it indicates the soccer ball landed below the initial height, meaning the height of the building is approximately 58.37 meters.

Would you like further details or have any questions?

Here are five related questions you might find interesting:

1. How would the time of flight change if the initial speed were increased?
2. What would happen if the angle of projection were 45 degrees?
3. How can the problem be adjusted if there was an initial height of the roof above the ball's launch point?
4. How would the height change if the horizontal distance was doubled?
5. What if the building height were known, and you needed to find the landing distance instead?

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