91Ó°ÊÓ

In C. S. Forster's novel Lieutenant Hornblower (set in the early \(1800 \mathrm{~s}\) ), a British naval vessel tries to sneak by a Spanish garrison. The ship passes as far away from the Spanish guns as it can -a distance \(s\). The Spanish gunner knows that his gun has a muzzle velocity whose magnitude is equal to \(v_{1}\). (a) Once the gun is fired, what controls the motion of the cannonball? Write the equations that determine the vector position of the cannonball after it leaves the cannon. You may ignore air resistance. (b) Suppose the gunner inclines his gun upward at an angle \(\theta\) to the horizontal. Solve the equations you have written in part (a) to obtain expressions that can be evaluated to give the position of the cannonball at any time, \(t\). (c) If the gunner wants the cannonball to hit the ship, he must choose his angle correctly. Explain how he can calculate the correct angle. (Again, you may ignore air resistance.) (d) If the muzzle velocity of the cannonball has a magnitude of \(100 \mathrm{~m} / \mathrm{s}\) and the ship is a distance of half a kilometer away, find the angle the gunner should use. (Take \(g\) to be \(10 \mathrm{~m} / \mathrm{s}^{2}\).)

Step by step solution

01

Understanding the problem and motion control

Identify what controls the motion of the cannonball. The motion of the cannonball after leaving the cannon is governed by the initial velocity vector and the forces of gravity. In the absence of air resistance, the only force acting on the cannonball is gravity.

02

Equations of motion

Write the equations for the vector position of the cannonball. The horizontal and vertical positions as functions of time are given by: \( x(t) = v_1 \cos\theta \cdot t \) \( y(t) = v_1 \sin\theta \cdot t - \frac{1}{2} gt^2 \)

03

Solving for the positions

Let's solve the equations to find the positions of the cannonball at a given time: \( x(t) = v_1 \cos\theta \cdot t \) \( y(t) = v_1 \sin\theta \cdot t - \frac{1}{2} gt^2 \)

04

Calculating the correct angle

To hit the ship at distance \( s \), the cannonball must travel a horizontal distance \( s \) in time \( t \). Set \( x(t) = s \): \( s = v_1 \cos\theta \cdot t \), Solve for \( t \): \( t = \frac{s}{v_1 \cos\theta} \).Plug \( t \) back into \( y(t) = 0 \) (to ensure the ball hits at the ship's height):\( 0 = v_1 \sin\theta \cdot \frac{s}{v_1 \cos\theta} - \frac{1}{2} g\left( \frac{s}{v_1 \cos\theta} \right)^2 \).

05

Solving for \( \theta \)

Simplify the equation to solve for the angle \( \theta \): \( 0 = \tan\theta \cdot s - \frac{1}{2} g \left( \frac{s^2}{v_1^2 \cos^2\theta } \right) \). Which leads to: \( 0 = \tan\theta \cdot v_1^2 - \frac{1}{2} gs^2 \sec^2\theta \) Solve for \( \theta \).

06

Specific Values Calculation

For the given values: \( v_1 = 100 \mathrm{m/s} \) \( s = 500\mathrm{m} \) \( g=10 \mathrm{m/s^2} \): Set up the equation: \( 500\mathrm{m} = 100 \cos\theta \cdot t \) Solve for \( t \): \( t = 5 / \cos\theta \) The height equation becomes: \( 0 = 100 \sin\theta \cdot 5 / \cos\theta - 0.5 \cdot 10 \left( 5 / \cos\theta \right)^2 \).

07

Final Angle Calculation

\( 0 = 500 \tan\theta - 2500 / \cos^2\theta \cis^2 \frac{ sec^2\theta = 1 + \tan^2\theta } \) yields: \( 0 = 500 \tan\theta - 2500(1 + \tan^2\theta) \). Solve for \( \theta \).

08

Result

\( \theta = \tan^{-1} \frac{ sgt\theta \) Results \( \theta = 26.56\theta \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinematic Equations

Understanding projectile motion begins with the kinematic equations. These equations describe the motion of an object under the influence of forces. For projectile motion without air resistance, gravity is the only force acting on the object after it is launched. Using these equations, you can determine the object's position at any given time. The horizontal position, denoted as \(x(t)\), is described by: \(x(t) = v_1 \cos\theta \cdot t\), where:

• \(v_1\) is the initial velocity (muzzle velocity)
• \(\theta\) is the angle of launch
• \(t\) is time in seconds

The vertical position, denoted as \(y(t)\), is given by: \(y(t) = v_1 \sin \theta \cdot t - \frac{1}{2} gt^2\), where:

• \(g\) is the acceleration due to gravity (9.8 \(m/s^2\), or 10 \(m/s^2\) for simplification)

Using these equations, we can determine an object's trajectory.

Projectile Range

Projectile range is the horizontal distance a projectile travels. It depends on the initial velocity and the launch angle. To find your projectile's range, we need to identify the time \(t\) at which the projectile hits the ground, which corresponds to \(y(t) = 0\). From the previously stated equations, setting \(x(t) = s\) lets us solve for the time \(t\): \(s = v_1 \cos\theta \cdot t\). Solving for time, we get \(t = \frac{s}{v_1 \cos\theta}\). Substituting \(t\) back in the vertical position equation, we make \(y(t) = 0\), resulting in: \(0 = v_1 \sin \theta \cdot \frac{s}{v_1 \cos \theta} - \frac{1}{2} g \left( \frac{s}{v_1 \cos \theta} \right)^2\). Solving this, the range formula can further optimize our launch angle calculations.

Angle of Launch

The angle at which a projectile is launched significantly affects its range and height. For the cannonball to hit the ship, we must carefully calculate the launch angle \(\theta\). Using the formulas from the kinematic equations, setting \(y(t) = 0\) to guarantee a hit, and solving \(0 = \tan\theta \cdot s - \frac{1}{2} g \left( \frac{s^2}{v_1^2 \sec^2\theta } \right)\), simplifies to: \(0 = \tan\theta \cdot v_1^2 - \frac{1}{2} g\tilde\cdot s^2 \left(1 + \tan^2\theta \right)\). Isolating \( \theta \) from here requires algebraic manipulation and the use of trigonometric identities.

Muzzle Velocity

Muzzle velocity refers to the initial speed of the projectile upon leaving the cannon. Denoted as \(v_1\), it is crucial for calculating both the range and the optimal launch angles. For instance, in our exercise, the given muzzle velocity is \(100 \ m/s\). We often resolve it into horizontal (\(v_1 \cos\theta\)) and vertical (\(v_1 \sin\theta\)) components. These provide us the basis for other critical calculations.

Gravity

Gravity is the force causing the projectile to follow a curved trajectory and eventually land. For simpler calculations, we often approximate gravity as \(10 \ m/s^2\) instead of \(9.8 \ m/s^2\). The acceleration due to gravity sets the parabolic path and influences when the projectile will hit the ground. The presence of \(g\) in equations like \(y(t) = v_1 \sin\theta \cdot t - \frac{1}{2} gt^2\) determines how quickly the projectile falls, essential to calculating collision with targets.