A cannonball is fired at v0 = 100 m/s from a barrel tilted upward at θ0 = 25 ∘. Find an expression for the cannonball's direction of travel, measured as an angle from horizontal, after traveling horizontal distance  d .

We are tasked with finding the direction of travel (i.e., the angle \(\theta(d)\)) of a cannonball as it moves after traveling a horizontal distance \(d\). This involves projectile motion, with an initial velocity \(v_0 = 100 \, \text{m/s}\) and a launch angle \(\theta_0 = 25^\circ\).

### Step 1: Components of initial velocity

The initial velocity can be broken down into horizontal and vertical components:
\[
v_{0x} = v_0 \cos(\theta_0)
\]
\[
v_{0y} = v_0 \sin(\theta_0)
\]

Substitute the known values:
\[
v_{0x} = 100 \cdot \cos(25^\circ)
\]
\[
v_{0y} = 100 \cdot \sin(25^\circ)
\]

### Step 2: Horizontal and vertical velocity after time \(t\)
The horizontal velocity remains constant because there is no horizontal acceleration (neglecting air resistance):
\[
v_x = v_{0x}
\]

The vertical velocity changes due to gravitational acceleration \(g = 9.8 \, \text{m/s}^2\):
\[
v_y(t) = v_{0y} - g t
\]

### Step 3: Time as a function of horizontal distance \(d\)
The horizontal distance \(d\) is related to time \(t\) by:
\[
d = v_x t
\]
Thus, the time is:
\[
t = \frac{d}{v_x}
\]

### Step 4: Vertical velocity in terms of \(d\)
Substitute the expression for \(t\) into the equation for vertical velocity:
\[
v_y(d) = v_{0y} - g \left( \frac{d}{v_x} \right)
\]

### Step 5: Angle of direction of travel
The direction of travel is the angle \(\theta(d)\) from the horizontal, which can be found using the arctangent of the ratio of vertical to horizontal velocity:
\[
\theta(d) = \tan^{-1}\left( \frac{v_y(d)}{v_x} \right)
\]

Substitute the expression for \(v_y(d)\):
\[
\theta(d) = \tan^{-1}\left( \frac{v_{0y} - g \left( \frac{d}{v_x} \right)}{v_x} \right)
\]

### Step 6: Final expression
Substitute the values for \(v_{0x}\), \(v_{0y}\), and \(v_x\) in terms of the initial velocity \(v_0\) and launch angle \(\theta_0\):
\[
\theta(d) = \tan^{-1}\left( \frac{v_0 \sin(\theta_0) - g \left( \frac{d}{v_0 \cos(\theta_0)} \right)}{v_0 \cos(\theta_0)} \right)
\]

This is the expression for the direction of travel as a function of the horizontal distance \(d\).

### Summary of the final formula:
\[
\theta(d) = \tan^{-1} \left( \frac{v_0 \sin(\theta_0) - \frac{g d}{v_0 \cos(\theta_0)}}{v_0 \cos(\theta_0)} \right)
\]

Would you like to see a worked example or have any other questions?

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Here are some related questions to deepen your understanding:
1. How does changing the initial velocity \(v_0\) affect the angle \(\theta(d)\)?
2. What is the angle \(\theta(d)\) at the maximum height of the projectile's trajectory?
3. How does the launch angle \(\theta_0\) influence the horizontal range of the cannonball?
4. What would happen to the trajectory if air resistance were considered?
5. How long does the cannonball stay in the air for a given horizontal distance \(d\)?

**Tip:** The maximum range of a projectile is achieved when the launch angle is \(45^\circ\), assuming no air resistance.

A cannonball is fired at v0 = 100 m/s from a barrel tilted upward at θ0 = 25 ∘. Find an expression for the cannonball's direction of travel, measured as an angle from horizontal, after traveling horizontal distance d.

Explore how to find the direction of travel of a cannonball after it covers a horizontal distance. This problem involves projectile motion, initial velocity decomposition, and trigonometric functions. Learn key kinematics concepts and formulas, suitable for students in Grades 11-12.

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