Problem 40 - Classical Mechanics

Question

A cannon shoots a ball at an angle \(\theta\) above the horizontal ground. (a) Neglecting air resistance, use Newton's second law to find the ball's position as a function of time. (Use axes with \(x\) measured horizontally and \(y\) vertically.) (b) Let \(r(t)\) denote the ball's distance from the cannon. What is the largest possible value of \(\theta\) if \(r(t)\) is to increase throughout the ball's flight? [Hint: Using your solution to part (a) you can write down \(r^{2}\) as \(x^{2}+y^{2},\) and then find the condition that \(r^{2}\) is always increasing.]

Step-by-Step Solution

Answer

\(\theta\) must be less than 45° for \(r(t)\) to always increase.

Step 1: Break Down the Problem

The exercise involves two main parts: deriving an expression for the ball's position using Newton's second law and then determining the condition under which the ball's distance from the cannon continuously increases. We'll first focus on finding the position as a function of time.

Step 2: Derive Position as a Function of Time

Consider the initial velocity of the ball as \( v_0 \) and it is launched at an angle \( \theta \). The initial velocity components can be written as \( v_{0x} = v_0 \cos(\theta) \) and \( v_{0y} = v_0 \sin(\theta) \). Using Newton's second law and neglecting air resistance, the only force acting on the ball is gravity, which affects only the vertical component (\( y \)). The equations of motion become: \[ x(t) = v_0 \cos(\theta) \cdot t \] \[ y(t) = v_0 \sin(\theta) \cdot t - \frac{1}{2}gt^2 \] where \( g \) is the acceleration due to gravity.

Step 3: Express Distance as a Function of Time

The distance \( r(t) \) from the origin (cannon) can be expressed using Pythagoras' theorem: \[ r(t) = \sqrt{ x(t)^2 + y(t)^2 } \] Thus, \[ r^2(t) = (v_0 \cos(\theta) \cdot t)^2 + \left(v_0 \sin(\theta) \cdot t - \frac{1}{2}gt^2 \right)^2 \] Expanding this, we need to find the condition to ensure \( r^2(t) \) is always increasing.

Step 4: Find Condition for Increasing \( r^2(t) \)

Differentiate \( r^2(t) \) with respect to \( t \) and set it greater than zero for it to always increase: \[ \frac{d}{dt} \Big( (v_0 \cos(\theta) \cdot t)^2 + \left(v_0 \sin(\theta) \cdot t - \frac{1}{2}gt^2 \right)^2 \Big) > 0 \] After simplifying, this leads to the condition: \[ 2v_0^2 \sin(\theta) \cos(\theta) > gt \] The inequality must hold true for the entire flight. Simplifying gives:\[ \sin(2\theta) > \frac{gt}{v_0} \]

Step 5: Determine the Largest Possible Angle \( \theta \)

The inequality \( \sin(2\theta) \) varies from \( 0 \) to \( 1 \), reaching its maximum when \( 2\theta = 90^\circ \) or \( \theta = 45^\circ \). Therefore, for the distance \( r(t) \) to continuously increase, the angle \( \theta \) must be less than \( 45^\circ \).

Key Concepts

Newton's Second Law

Newton's Second Law is foundational in understanding projectile motion. It states that the acceleration of an object depends on two variables: the net force acting upon the object and the mass of the object. In our exercise, the cannonball's motion can be analyzed using this law.
Gravity is the only force acting on the cannonball, pulling it downwards. This means the horizontal motion is constant, while the vertical motion is accelerated by gravity.
To find the ball's position, we consider:

• The horizontal component with no acceleration: \( x(t) = v_0 \cos(\theta) \cdot t \)
• The vertical component with acceleration from gravity: \( y(t) = v_0 \sin(\theta) \cdot t - \frac{1}{2}gt^2 \)

By using these equations, we capture how the ball moves through time in both the horizontal and vertical directions.

Parabolic Trajectory

Projectiles, like cannonballs, often follow a curved path known as a parabolic trajectory. This shape emerges due to the constant horizontal velocity and the accelerating vertical motion influenced by gravity.
When you throw or shoot something at an angle, both these motions come together, leading to the classic parabola shape.
Key points about parabolic trajectories:

• The highest point, or apex, occurs when the vertical velocity component is zero. At this point, the projectile stops rising and starts falling.
• The symmetry of the path means the descent mirrors the ascent.

Understanding this helps in predicting where and when the projectile will land, crucial for tasks like determining how far a cannonball will travel.

Optimal Launch Angle

The angle at which you launch a projectile can dramatically affect how far it travels. The optimal launch angle is the angle that provides the greatest range for a given initial speed.
For projectiles launched with the intent of maximizing distance, 45 degrees usually reigns supreme. This angle balances the vertical and horizontal components, optimizing both height and range.
However, in our exercise, the condition differs slightly. To ensure the ball’s distance from the cannon always increases:

• The launch angle \( \theta \) should be less than 45 degrees.
• This ensures the derivative of the squared distance is positive, avoiding the situation where the projectile starts falling back toward the cannon.

Choosing the correct launch angle is vital for controlling and predicting projectile behavior effectively.