Given:

• Initial speed ($v_0$) = 14 m/s
• Launch angle ($\theta$) = 60 degrees
• Height of the building ($h$) = 4 m
• Gravity ($g$) = 9.81 m/s²

We will break the problem into parts and solve each step by step.

Part 1: Final vertical displacement ($\Delta y$)

The final vertical displacement is the difference in height between the landing point (the roof) and the launch point (ground level). Since the ball lands on the roof which is 4 meters above the launch point, the final vertical displacement is:

$\Delta y = 4 \, \text{m}$

Part 2: Time of Flight ($t$)

We can find the time of flight using the vertical motion equation:

$y = v_{0y} t - \frac{1}{2} g t^2$

Where:

• $y$ is the vertical displacement (which is 4 m).
• $v_{0y} = v_0 \sin(\theta)$ is the initial vertical velocity.
• $g$ is the acceleration due to gravity.

Let’s calculate $v_{0y}$:

$v_{0y} = 14 \times \sin(60^\circ) = 14 \times \frac{\sqrt{3}}{2} \approx 12.12 \, \text{m/s}$

Now we substitute into the equation:

$4 = 12.12 t - \frac{1}{2} \times 9.81 \times t^2$

This simplifies to:

$4 = 12.12 t - 4.905 t^2$

Rearrange into standard quadratic form:

$4.905 t^2 - 12.12 t + 4 = 0$

Now, we can solve this quadratic equation using the quadratic formula:

$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Where:

• $a = 4.905$
• $b = -12.12$
• $c = 4$

Let me compute the values.

The two solutions for the time of flight are:

• $t_1 = 2.08 \, \text{seconds}$
• $t_2 = 0.39 \, \text{seconds}$

Since $t_2$ corresponds to an earlier point (during the upward trajectory), we discard it, and the valid time of flight is $t = 2.08 \, \text{seconds}$.

Part 3: Horizontal Displacement ($x$)

The horizontal displacement can be calculated using the horizontal component of the velocity and the time of flight. The horizontal velocity $v_{0x}$ is:

$v_{0x} = v_0 \cos(\theta)$

Let’s calculate $v_{0x}$:

$v_{0x} = 14 \times \cos(60^\circ) = 14 \times 0.5 = 7 \, \text{m/s}$

Now, the horizontal displacement is:

$x = v_{0x} \times t = 7 \times 2.08 = 14.56 \, \text{m}$

Summary:

• Final vertical displacement: $4 \, \text{m}$
• Time of flight: $2.08 \, \text{seconds}$
• Horizontal displacement: $14.56 \, \text{m}$

Would you like more details on any part of the solution?

Related Questions:

1. How would the time of flight change if the launch speed was higher?
2. What would be the horizontal displacement if the angle of launch was different?
3. How would the trajectory change if air resistance were considered?
4. What happens if the building height increases to 6 meters?
5. How would you find the maximum height reached by the ball during its flight?

Tip:

Always break down projectile motion into horizontal and vertical components for simpler calculations!