A projectile of mass 0.741 kg is shot from a
cannon. The end of the cannon’s barrel is
at height 7.1 m, as shown in the figure. The
initial velocity of the projectile is 11 m/s.
The projectile rises to a maximum height of
∆y above the end of the cannon’s barrel and
strikes the ground a horizontal distance ∆x
past the end of the cannon’s barrel.Find the range ∆x of the projectile. The
acceleration of gravity is 9.8 m/s
Answer in units of m.

To find the range (\(\Delta x\)) of the projectile, we can apply the standard equations of projectile motion.

### Step 1: Resolve the Initial Velocity into Components
The initial velocity of the projectile is given as \( v_0 = 11 \, \text{m/s} \).

Assuming the angle of projection is \( \theta \), the components of the initial velocity are:
- Horizontal velocity component: \( v_{0x} = v_0 \cos \theta \)
- Vertical velocity component: \( v_{0y} = v_0 \sin \theta \)

Since the angle \( \theta \) is not provided, I will need clarification from you on that, or we can calculate it if given other conditions.

### Step 2: Determine Time of Flight
To find the time of flight \( t \), we use the vertical motion equations.

The vertical displacement is given by:
\[
y = y_0 + v_{0y}t - \frac{1}{2}gt^2
\]
Where:
- \( y_0 = 7.1 \, \text{m} \) (initial height)
- \( y = 0 \, \text{m} \) (the projectile hits the ground)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)

Substituting these values, the equation becomes:
\[
0 = 7.1 + v_{0y}t - \frac{1}{2}(9.8)t^2
\]

We will need the value of \( v_{0y} \), which depends on the angle \( \theta \).

### Step 3: Calculate the Range
The range \( \Delta x \) is given by:
\[
\Delta x = v_{0x} \cdot t
\]
Where \( v_{0x} \) is the horizontal velocity component and \( t \) is the time of flight.

To proceed with the calculation, I need either the angle \( \theta \) or more details on the problem setup. Could you provide that, or would you like me to assume a certain value for the angle?

---

Let me know if you would like more details or have questions about any step. Here are five related questions for further exploration:

1. How does the angle of projection affect the range of a projectile?
2. What happens to the maximum height if the initial speed is increased?
3. How does the height of the barrel affect the total time of flight?
4. Can air resistance significantly alter the range of the projectile in real-world conditions?
5. What is the relationship between horizontal and vertical components in projectile motion?

**Tip:** In projectile motion, the horizontal velocity remains constant throughout the flight, while vertical velocity changes due to gravity.

A projectile of mass 0.741 kg is shot from a cannon. The end of the cannon’s barrel is at height 7.1 m. The initial velocity of the projectile is 11 m/s. The projectile rises to a maximum height above the end of the cannon’s barrel and strikes the ground a horizontal distance ∆x past the end of the cannon’s barrel. Find the range ∆x of the projectile. The acceleration of gravity is 9.8 m/s². Answer in units of meters.

Solve a projectile motion problem where a projectile is shot from a cannon at a height of 7.1 m with an initial velocity of 11 m/s. Learn how to calculate the horizontal range ∆x of the projectile, covering kinematics and 2D motion concepts. Suitable for Grade 10-12 students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation