An artillery shell is fired with an initial velocity of 300 m/s at 54.0° above the horizontal. To clear an avalanche, it explodes on a mountainside 40.5 s after firing. What are the x- and y-coordinates of the shell where it explodes, relative to its firing point?

To solve for the \( x \)- and \( y \)-coordinates of the shell at the point of explosion, we need to break down the motion into its horizontal (\( x \)) and vertical (\( y \)) components using the initial velocity and the time of flight.

### Given:
- Initial velocity (\( v_0 \)) = 300 m/s
- Launch angle (\( \theta \)) = 54.0°
- Time of flight (\( t \)) = 40.5 s
- Acceleration due to gravity (\( g \)) = 9.8 m/s²

### Step 1: Resolve initial velocity into horizontal and vertical components.
The initial velocity can be split into two components:
- Horizontal component: \( v_{0x} = v_0 \cos(\theta) \)
- Vertical component: \( v_{0y} = v_0 \sin(\theta) \)

\[
v_{0x} = 300 \cos(54^\circ)
\]
\[
v_{0y} = 300 \sin(54^\circ)
\]

### Step 2: Calculate the horizontal displacement (\( x \)-coordinate).
The horizontal displacement is given by:

\[
x = v_{0x} \cdot t
\]

Since there is no acceleration in the horizontal direction (ignoring air resistance), the horizontal velocity remains constant.

### Step 3: Calculate the vertical displacement (\( y \)-coordinate).
The vertical displacement is affected by gravity, and the equation for vertical displacement is:

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

Now, let's compute these components step by step. I'll calculate the values for you.The coordinates of the shell at the point where it explodes are:

- \( x \)-coordinate: 7141.6 meters
- \( y \)-coordinate: 1792.3 meters

These are the horizontal and vertical distances from the firing point at the moment of the explosion.

Would you like further details or have any questions?

Here are 5 related questions:
1. How would air resistance affect the projectile's trajectory?
2. How would a different launch angle impact the range and height?
3. What would be the maximum height the shell reaches during its flight?
4. How would the time of flight change if the launch velocity were different?
5. How do we calculate the impact velocity of the shell?

**Tip:** When solving projectile motion problems, it's helpful to break down the motion into horizontal and vertical components, treating each independently.

Explore the physics of projectile motion by calculating the x- and y-coordinates of an artillery shell fired with an initial velocity of 300 m/s at an angle of 54.0° above the horizontal. The shell explodes on a mountainside 40.5 seconds after launch. This problem involves breaking down the motion into horizontal and vertical components using kinematic equations. Suitable for high school students in Grades 10-12.

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