https://file.aiquickdraw.com/mathbot/file/1727187945337fmmuzmyk.jpg

This is a projectile motion problem. Let's break down the information provided and solve it step by step:

### Given:
- The height of the vertical wall (with railing): \( 7 + 1 = 8 \, \text{m} \) (since the 7m wall includes a 1m high railing).
- The ball is launched from a point \( 24 \, \text{m} \) away from the base of the wall at an angle of \( 53^\circ \).
- The time taken to reach the point directly above the wall is \( 2.2 \, \text{seconds} \).
- We are tasked to find the horizontal distance from the wall to where the ball lands on the roof.

### Step-by-Step Breakdown:

1. **Horizontal Component of Velocity**:
   The horizontal component of the velocity is constant throughout the motion. Let’s find it.

   \[
   v_x = v_0 \cos(\theta)
   \]
   where \( v_0 \) is the initial velocity and \( \theta = 53^\circ \).

2. **Vertical Component of Velocity**:
   The vertical component of velocity affects the time of flight to the point above the wall:

   \[
   v_y = v_0 \sin(\theta)
   \]

3. **Height Calculation**:
   The ball reaches a point vertically above the wall, so the height must account for the wall's height of \( 8 \, \text{m} \).

4. **Horizontal Distance**:
   The key is to calculate the distance it travels horizontally during the projectile's motion using the time and velocity.

Let me calculate everything for you.The initial velocity of the ball when it was launched is approximately \( 18.07 \, \text{m/s} \).

The horizontal component of the velocity is \( 10.87 \, \text{m/s} \).

However, it seems that the horizontal distance to the point directly above the wall is almost the same as the launch distance (24 m), leaving only a negligible distance of around \(-0.08 \, \text{m}\). This suggests that the ball lands very close to the vertical line of the wall, which means the ball just barely reaches the point above the wall on the roof.

Would you like more details or any clarifications?

Here are 5 related questions to explore:
1. How would the result change if the angle of launch was increased?
2. What would happen if the wall height was increased to 10 m?
3. How does air resistance affect projectile motion in real-world scenarios?
4. What is the relationship between launch speed and the maximum height a projectile can reach?
5. How would the time of flight change if the initial velocity was doubled?

**Tip:** In projectile motion, breaking velocity into horizontal and vertical components simplifies the calculations significantly.

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 d=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.

In this projectile motion problem, a ball is launched from a point 24m away at an angle of 53 degrees. Using the provided time and velocity components, learn how to calculate the horizontal distance the ball travels to land on the roof. This problem involves kinematics and is suitable for Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation