Question

2.16 \(\star\) A golfer hits his ball with speed \(v_{\mathrm{o}}\) at an angle \(\theta\) above the horizontal ground. Assuming that the angle \(\theta\) is fixed and that air resistance can be neglected, what is the minimum speed \(v_{\mathrm{o}}(\min )\) for which the ball will clear a wall of height \(h\), a distance \(d\) away? Your solution should get into trouble if the angle \(\tan \theta < h / d .\) Explain. What is \(v_{\mathrm{o}}(\min )\) if \(\theta=25^{\circ}, d=50 \mathrm{m},\) and \(h=2 \mathrm{m} ?\)

Short answer

\(v_0(\min) \approx 14.15\,\text{m/s}\) for \(\theta=25^{\circ}\), \(d=50\,\text{m}\), and \(h=2\,\text{m}\). If \(\tan \theta < \frac{h}{d}\), the wall cannot be cleared.

Step-by-step solution

Identify the Key Elements of Projectile Motion

The ball is launched with initial speed \(v_0\) at an angle \(\theta\) with the horizontal. The key equations of motion are the horizontal and vertical components of the velocity. The horizontal distance \(d\) is covered in time \(t\), and the vertical height \(h\) must be reached at that time. The horizontal component is \(v_0\cos\theta\) and the vertical component is \(v_0\sin\theta\).

Write the Equations of Motion

For horizontal motion: \[d = v_0 \cos \theta \cdot t\]For vertical motion:\[h = v_0 \sin \theta \cdot t - \frac{1}{2}gt^2\]where \(g\) is the acceleration due to gravity. Solve these simultaneously.

Solve for Time \(t\)

From the horizontal equation, solve for time:\[t = \frac{d}{v_0 \cos \theta}\]Substitute this expression for \(t\) into the vertical equation.

Substitute and Simplify the Vertical Motion Equation

Substitute \(t = \frac{d}{v_0 \cos \theta}\) into the vertical equation:\[h = v_0 \sin \theta \cdot \frac{d}{v_0 \cos \theta} - \frac{1}{2}g\left(\frac{d}{v_0 \cos \theta}\right)^2\]Simplify:\[h = d \tan \theta - \frac{g d^2}{2 v_0^2 \cos^2 \theta}\]

Rearrange for Minimum Speed \(v_0(\min)\)

Solve for \(v_0\):\[\v_0^2 = \frac{g d^2}{2(h - d \tan \theta) \cos^2 \theta}\]The minimum speed occurs when \(v_0 = v_0(\min)\):\[\v_0(\min) = \sqrt{\frac{g d^2}{2(h - d \tan \theta) \cos^2 \theta}}\]

Determine if Condition \(\tan \theta < \frac{h}{d}\) Occurs

The derived expression for \(v_0(\min)\) becomes undefined if \(h < d \tan \theta\). This means the ball can never achieve sufficient height to clear the wall if \(\tan \theta < \frac{h}{d}\). Verify this condition does not hold before proceeding.

Calculate the Minimum Speed for Given Values

Substitute \(\theta = 25^{\circ}, d = 50\,\text{m}, h = 2\,\text{m} \) into the formula:Convert \(\theta\) to radians for calculation, \(\theta = \frac{25\pi}{180}\).Calculate \(\tan(\theta) = \tan(25^{\circ})\) and \(\cos(25^{\circ})\).\Insert values:\[v_0(\min) = \sqrt{\frac{9.8 \cdot 50^2}{2(2 - 50 \cdot \tan(25^{\circ})) \cdot \cos^2(25^{\circ})}}\]Solve to find \(v_0(\min)\).

Key concepts

Initial Velocity Determination

Determining the initial velocity, denoted as \( v_0 \), is a critical step in solving projectile motion problems. The initial velocity is the speed at which an object is projected into the air at a specific angle \( \theta \). In this scenario with the golfer's ball, the aim is to find the minimum initial speed \( v_0(\min) \) necessary for the ball to clear a wall of height \( h \) placed at a distance \( d \).

This involves understanding the relationship between the launch angle, the initial speed, and the desired trajectory. If the angle \( \theta \) and distances \( h \) and \( d \) are provided, we can utilize these parameters to determine the minimum initial velocity required to achieve the projectile mission using kinematic equations. This forms the basis of addressing projectile-related queries. Ensure to verify initial conditions like \( \tan \theta \geq \frac{h}{d} \) to ensure a feasible trajectory.

Equations of Motion

The equations of motion are essential in understanding projectile trajectories. They help us break down the path of the object into horizontal and vertical components, which can be analyzed separately.

For horizontal motion, the equation is given by:
\[d = v_0 \cos \theta \cdot t\]
where,

• \( d \) is the horizontal distance, and
• \( v_0 \cos \theta \) represents the horizontal component of initial velocity.

Vertical motion can be expressed as:
\[h = v_0 \sin \theta \cdot t - \frac{1}{2}gt^2\]
where,

• \( h \) is the vertical height,
• \( g \) is the acceleration due to gravity, and
• \( v_0 \sin \theta \) is the vertical component of the initial speed.

These equations allow us to solve for time \( t \) and subsequently use it to determine other unknowns like the initial velocity.

Horizontal and Vertical Components

Projectile motion is broken into horizontal and vertical components to simplify the analysis. The horizontal component \( v_0 \cos \theta \) influences how far the projectile travels along the ground, while the vertical component \( v_0 \sin \theta \) influences how high it can go.

Let's break them down:

• Horizontal Component: Since there's no acceleration in the horizontal direction (neglecting air resistance), the horizontal component remains constant.
• Vertical Component: This component is affected by gravity. It determines the projectile's maximum height and the time it takes to reach that height.

Grasping these components helps us work out solutions effectively by simplifying the forces acting on the projectile into manageable parts. This simplification opens the way to calculate time \( t \) and how it affects the shooting conditions.

Minimum Speed Calculation

The minimum speed calculation for a projectile to clear an obstacle involves deriving a formula that calculates \( v_0(\min) \). This means finding the lowest speed at which the ball can meet the requirements of reaching a given height and distance.

The derived formula for \( v_0(\min) \) is:
\[v_0(\min) = \sqrt{\frac{g d^2}{2(h - d \tan \theta) \cos^2 \theta}}\]
Here, the steps are:

• Rearrange the vertical motion equation to express height \( h \) in terms of the horizontal distance \( d \), using \( \tan \theta \).
• Substitute and solve for \( v_0 \) by ensuring all variables meet the criteria provided.

Keep in mind, \( v_0(\min) \) becomes undefined if the condition \( \tan \theta < \frac{h}{d} \) is met. Hence, it’s a critical step to check if conditions make the solution feasible as you proceed with the calculations.