A playground is on the flat roof of a city school, 6m above the street below. the vertical wall of the building is 7m high, forming 1m high railing around the playground. A ball has fallen into the street below. A passerby returns it by launching the ball at an angle of 53 degrees above the horizontal at a point d=24m away from the base of the building. The ball takes 2.2 seconds to a reach a point vertically above the wall. Find the horizontal distance from the wall to the point on the roof where the ball lands.

To solve this problem, we need to analyze the motion of the ball in two dimensions: horizontal and vertical. Let's break it down:

### Step 1: Separate Horizontal and Vertical Components
The ball is launched at an angle of \( 53^\circ \) with a velocity \( v_0 \). We don't know the initial speed \( v_0 \), but we can resolve it into horizontal and vertical components:
- Horizontal velocity: \( v_{0x} = v_0 \cos 53^\circ \)
- Vertical velocity: \( v_{0y} = v_0 \sin 53^\circ \)

### Step 2: Vertical Motion Analysis
The ball takes \( 2.2 \, \text{seconds} \) to reach a point vertically above the wall, which is \( 7 \, \text{m} \) high. We will use the following kinematic equation for vertical motion:

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

Where:
- \( y = 7 \, \text{m} \) (the vertical displacement to the top of the wall)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( t = 2.2 \, \text{seconds} \) (time to reach above the wall)

Substitute these values into the equation:

\[
7 = v_0 \sin 53^\circ (2.2) - \frac{1}{2} (9.8) (2.2)^2
\]

### Step 3: Solve for \( v_0 \)
Simplify the equation to find the initial velocity \( v_0 \):

\[
7 = v_0 \sin 53^\circ (2.2) - \frac{1}{2} (9.8) (4.84)
\]
\[
7 = v_0 (0.7986)(2.2) - 23.716
\]
\[
7 + 23.716 = 1.757 v_0
\]
\[
v_0 = \frac{30.716}{1.757} \approx 17.48 \, \text{m/s}
\]

### Step 4: Horizontal Motion Analysis
Now that we have the initial speed, we can find how far the ball travels horizontally. The horizontal velocity is:

\[
v_{0x} = v_0 \cos 53^\circ = 17.48 \times 0.6018 \approx 10.51 \, \text{m/s}
\]

The horizontal distance traveled in \( t = 2.2 \, \text{seconds} \) is:

\[
x_{\text{wall}} = v_{0x} \times t = 10.51 \times 2.2 \approx 23.12 \, \text{m}
\]

### Step 5: Find the Remaining Horizontal Distance to the Roof
The passerby is \( d = 24 \, \text{m} \) away from the building, so the ball has traveled \( 23.12 \, \text{m} \) horizontally to reach a point directly above the wall. The remaining horizontal distance from the wall to where the ball lands on the roof is:

\[
x_{\text{roof}} = 24 - 23.12 \approx 0.88 \, \text{m}
\]

### Final Answer:
The horizontal distance from the wall to the point on the roof where the ball lands is approximately **0.88 meters**.

---

Would you like further details or clarifications? Here are some additional questions you might want to explore:

1. How would changing the launch angle affect the landing position?
2. What would happen if the ball were launched with a different initial speed?
3. Can you calculate the maximum height the ball reaches during its trajectory?
4. How would air resistance alter the results of this problem?
5. What is the total horizontal distance the ball travels?

**Tip**: In projectile motion, the horizontal and vertical components of motion are independent, so you can solve them separately using appropriate kinematic equations.

A playground is on the flat roof of a city school, 6m above the street below. The vertical wall of the building is 7m high, forming a 1m high railing around the playground. A ball has fallen into the street below. A passerby returns it by launching the ball at an angle of 53 degrees above the horizontal at a point 24m away from the base of the building. The ball takes 2.2 seconds to reach a point vertically above the wall. Find the horizontal distance from the wall to the point on the roof where the ball lands.

This detailed solution covers a projectile motion problem where a ball is launched at 53 degrees from 24m away to reach a rooftop 6m above the street. Learn how to calculate the initial velocity and the horizontal distance from the wall to where the ball lands on the roof using kinematic equations. Suitable for Grades 11-12.

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