Kayla throws a soccer ball out of her dorm window to Hafsa, who is waiting below to catch it. Kayla throws the ball at an angle of 40° below horizontal with a speed of 2.2 m/s, and Hafsa catches it 2.6 s later. How far from the base of the dorm does Hafsa stand to catch the ball, in meters? Use g = 10.0 m/s².

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### Projectile Motion Problem

**Problem Statement:**

Kayla throws a soccer ball out of her dorm window to Hafsa, who is waiting below to catch it. Kayla throws the ball at an angle of 40° below horizontal with a speed of 2.2 m/s, and Hafsa catches it 2.6 seconds later. How far from the base of the dorm does Hafsa stand to catch the ball, in meters? Use \( g = 10.0 \, \text{m/s}^2 \).

**Explanation:**

This problem illustrates the principles of projectile motion where an object is thrown with an initial velocity at an angle to the horizontal axis and is under the influence of gravity. Here, you have the initial speed, the angle of projection, and the time of flight, and you need to calculate the horizontal displacement (range) of the ball.

**Given Data:**
- Initial speed (u): 2.2 m/s
- Angle of projection (θ): 40° below horizontal
- Time of flight (t): 2.6 s
- Acceleration due to gravity (g): 10.0 m/s²

**Objective:**
Calculate the horizontal distance from the base of the dorm to where Hafsa catches the ball.

**Steps to Solve:**
1. Resolve the initial velocity into horizontal and vertical components using the given angle:
   - \( u_x = u \cdot \cos(\theta) \)
   - \( u_y = u \cdot \sin(\theta) \)

2. Horizontal distance covered in projectile motion is given by:
   - \( \text{Distance} = u_x \cdot t \)

**Detailed Calculation:**
1. Resolve the velocities:
   - \( u_x = 2.2 \cdot \cos(40°) \approx 1.685 \, \text{m/s} \)
   - \( u_y = 2.2 \cdot \sin(40°) \approx 1.414 \, \text{m/s} \) (below horizontal)

2. Calculate the horizontal distance:
   - \( \text{Distance} = u_x \cdot t \)
   - \( \text{Distance} = 1.685 \, \text{m/s} \cdot 2.6 \, \text{s} \approx 4.381 \, \
Transcribed Image Text:### Projectile Motion Problem **Problem Statement:** Kayla throws a soccer ball out of her dorm window to Hafsa, who is waiting below to catch it. Kayla throws the ball at an angle of 40° below horizontal with a speed of 2.2 m/s, and Hafsa catches it 2.6 seconds later. How far from the base of the dorm does Hafsa stand to catch the ball, in meters? Use \( g = 10.0 \, \text{m/s}^2 \). **Explanation:** This problem illustrates the principles of projectile motion where an object is thrown with an initial velocity at an angle to the horizontal axis and is under the influence of gravity. Here, you have the initial speed, the angle of projection, and the time of flight, and you need to calculate the horizontal displacement (range) of the ball. **Given Data:** - Initial speed (u): 2.2 m/s - Angle of projection (θ): 40° below horizontal - Time of flight (t): 2.6 s - Acceleration due to gravity (g): 10.0 m/s² **Objective:** Calculate the horizontal distance from the base of the dorm to where Hafsa catches the ball. **Steps to Solve:** 1. Resolve the initial velocity into horizontal and vertical components using the given angle: - \( u_x = u \cdot \cos(\theta) \) - \( u_y = u \cdot \sin(\theta) \) 2. Horizontal distance covered in projectile motion is given by: - \( \text{Distance} = u_x \cdot t \) **Detailed Calculation:** 1. Resolve the velocities: - \( u_x = 2.2 \cdot \cos(40°) \approx 1.685 \, \text{m/s} \) - \( u_y = 2.2 \cdot \sin(40°) \approx 1.414 \, \text{m/s} \) (below horizontal) 2. Calculate the horizontal distance: - \( \text{Distance} = u_x \cdot t \) - \( \text{Distance} = 1.685 \, \text{m/s} \cdot 2.6 \, \text{s} \approx 4.381 \, \
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