Projectile Motion I A projectile is launched at the angle 35° from the horizontal with velocity equal to 30 m/s. Neglecting air resistance and assuming a horizontal surface, determine how far away from the launch site the projectile will land. Figure 4.26 Projectile motion. To answer this problem, we will need: 1. Excel's trigonometry functions to handle the 35° angle, and; 2. Equations relating distance to velocity and acceleration.
Projectile Motion I A projectile is launched at the angle 35° from the horizontal with velocity equal to 30 m/s. Neglecting air resistance and assuming a horizontal surface, determine how far away from the launch site the projectile will land. Figure 4.26 Projectile motion. To answer this problem, we will need: 1. Excel's trigonometry functions to handle the 35° angle, and; 2. Equations relating distance to velocity and acceleration.
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![### Projectile Motion I
**Problem Statement:**
A projectile is launched at an angle of 35° from the horizontal with a velocity of 30 m/s. Neglecting air resistance and assuming a horizontal surface, determine how far away from the launch site the projectile will land.
**Figure 4.26:**
- This figure depicts the motion of a projectile. The projectile is shown being launched at a 35° angle from the horizontal.
- The path of the projectile follows a parabolic trajectory.
- The diagram includes labels indicating the angles and the horizontal distance, x, from the launch site to the point where the projectile will land.
**To answer this problem, we will need:**
1. **Excel’s trigonometry functions** to handle the 35° angle, and;
2. **Equations relating distance to velocity and acceleration.**
---
**Discussion:**
This problem can be approached using the principles of projectile motion. The key to solving it is to decompose the initial velocity into horizontal and vertical components:
- **Horizontal Component (Vx):**
\[
Vx = V \cdot \cos(\theta)
\]
where \( V \) is the initial velocity (30 m/s), and \( \theta \) is the launch angle (35°).
- **Vertical Component (Vy):**
\[
Vy = V \cdot \sin(\theta)
\]
The time of flight (T) of the projectile can be found by analyzing the vertical motion, considering the acceleration due to gravity (g = 9.81 m/s²).
Using this time of flight, the horizontal distance (range) can be calculated.
**Steps:**
1. Calculate the horizontal and vertical components of the initial velocity.
2. Determine the time of flight using the vertical motion equations.
3. Calculate the horizontal range using the time of flight and the horizontal velocity.
These calculations involve the use of standard kinematic equations, often simplified by breaking the motion into horizontal and vertical components. Excel functions can be utilized to handle the trigonometric calculations accurately.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe42c0efe-14fe-42af-bc89-aaae6e3d971b%2F3b8b4bd4-d273-4be9-b29f-901e55d8bad8%2Fubbcdnz_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Projectile Motion I
**Problem Statement:**
A projectile is launched at an angle of 35° from the horizontal with a velocity of 30 m/s. Neglecting air resistance and assuming a horizontal surface, determine how far away from the launch site the projectile will land.
**Figure 4.26:**
- This figure depicts the motion of a projectile. The projectile is shown being launched at a 35° angle from the horizontal.
- The path of the projectile follows a parabolic trajectory.
- The diagram includes labels indicating the angles and the horizontal distance, x, from the launch site to the point where the projectile will land.
**To answer this problem, we will need:**
1. **Excel’s trigonometry functions** to handle the 35° angle, and;
2. **Equations relating distance to velocity and acceleration.**
---
**Discussion:**
This problem can be approached using the principles of projectile motion. The key to solving it is to decompose the initial velocity into horizontal and vertical components:
- **Horizontal Component (Vx):**
\[
Vx = V \cdot \cos(\theta)
\]
where \( V \) is the initial velocity (30 m/s), and \( \theta \) is the launch angle (35°).
- **Vertical Component (Vy):**
\[
Vy = V \cdot \sin(\theta)
\]
The time of flight (T) of the projectile can be found by analyzing the vertical motion, considering the acceleration due to gravity (g = 9.81 m/s²).
Using this time of flight, the horizontal distance (range) can be calculated.
**Steps:**
1. Calculate the horizontal and vertical components of the initial velocity.
2. Determine the time of flight using the vertical motion equations.
3. Calculate the horizontal range using the time of flight and the horizontal velocity.
These calculations involve the use of standard kinematic equations, often simplified by breaking the motion into horizontal and vertical components. Excel functions can be utilized to handle the trigonometric calculations accurately.
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