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.