Problem 48 - College Physics

Question

An errand of mercy. An airplane is dropping bales of hay to cattle stranded in a blizzard on the Great Plains. The pilot releases the bales at 150 \(\mathrm{m}\) above the level ground when the plane is flying at 75 \(\mathrm{m} / \mathrm{s} 55^{\circ}\) above the horizontal. How far in front of the cattle should the pilot release the hay so that the bales will land at the point where the cattle are stranded?

Step-by-Step Solution

Answer

Release the bales approximately 312.5 meters ahead of the cattle.

Step 1: Identify known values

Firstly, identify the values given in the problem that we will use in our calculations. We know the following:- Height of the release point above ground, \( h = 150 \) m.- Initial speed of the plane, \( v_0 = 75 \) m/s.- Angle of release above the horizontal, \( \theta = 55^{\circ} \).We need to determine the horizontal distance the bales will travel before hitting the ground.

Step 2: Resolve initial velocity into components

Calculate the initial velocity components:- Horizontal component (\( v_{0x} \)): \[ v_{0x} = v_0 \cdot \cos(\theta) = 75 \cdot \cos(55^{\circ}) \]- Vertical component (\( v_{0y} \)): \[ v_{0y} = v_0 \cdot \sin(\theta) = 75 \cdot \sin(55^{\circ}) \]This step helps us understand the motion separately in the horizontal and vertical directions.

Step 3: Calculate time of flight using vertical motion

The time it takes for the bales to hit the ground is determined using the vertical motion equations. We use the equation,\[ h = v_{0y} \cdot t + \frac{1}{2}gt^2 \]where \( g = -9.8 \; m/s^2 \) is the acceleration due to gravity. Rearrange and solve for \( t \), the time of flight:\[ 0 = 150 + v_{0y}t - \frac{1}{2} \cdot 9.8 \cdot t^2 \]

Step 4: Solve for time of flight

Plug in the value of \( v_{0y} \) from Step 2 and solve the quadratic equation for \( t \). This will usually involve rearranging into a form \( at^2 + bt + c = 0 \) and solving using the quadratic formula,\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Step 5: Calculate horizontal distance

With the time of flight \( t \) calculated, determine the horizontal distance \( x \) traveled using:\[ x = v_{0x} \cdot t \]where \( v_{0x} \) is the horizontal component of the velocity found in Step 2.

Step 6: Final calculation and interpretation

After calculating \( x \), this will be the horizontal distance from the release point to the point of impact on the ground where the cattle are located. Make sure all values are appropriately calculated for real-world application, taking into account any assumptions like air resistance.

Key Concepts

Horizontal Distance Calculation

In projectile motion, the horizontal distance traveled by an object is crucial for determining where it will land. Understanding this helps us solve real-world problems, like when a pilot needs to drop supplies accurately. The horizontal distance, also known as the range, can be calculated once the time of flight and the horizontal component of the initial velocity are known. This is given by the formula:

• Horizontal Distance (x): \( x = v_{0x} \cdot t \)

Here, \( v_{0x} \) is the horizontal component of the initial velocity, and \( t \) is the time the object is in the air. This formula assumes there is no air resistance affecting the horizontal motion. By accurately computing the horizontal distance, we ensure that the materials fall exactly where needed. This is particularly critical in situations like delivering supplies to stranded cattle during a blizzard.

Initial Velocity Components

To understand the motion of a projectile, we need to break down its initial velocity into two parts: horizontal and vertical components.

• The Horizontal Component: \( v_{0x} = v_0 \cdot \cos(\theta) \)
• The Vertical Component: \( v_{0y} = v_0 \cdot \sin(\theta) \)

In this context, \( v_0 \) represents the initial speed of the object, and \( \theta \) is the angle of release. These components are important because they allow us to analyze the motion separately in the horizontal and vertical directions.

• The horizontal component, \( v_{0x} \), remains constant if we ignore air resistance.
• The vertical component, \( v_{0y} \), is affected by gravity, causing the object to accelerate downward.

By resolving the initial velocity into these components, it helps predict the projectile's path and landing location efficiently. This method is particularly useful in calculating the release point of the hay bales from a moving airplane.

Quadratic Equation for Time of Flight

The time of flight of a projectile is a vital factor in determining how far it will travel. For projectiles launched at an angle, this is found by analyzing their vertical motion.We use the equation:\[ h = v_{0y} \cdot t + \frac{1}{2}gt^2 \]In this equation, \( h \) is the vertical height from which the projectile is released, \( v_{0y} \) is the initial vertical velocity, \( g \) is the acceleration due to gravity (approximately \(-9.8 \; m/s^2\)), and \( t \) is the time of flight. This setup leads to a quadratic equation, which we need to solve for \( t \).Rearranging it, we find:\[ 0 = h + v_{0y}t - \frac{1}{2}gt^2 \]Using the quadratic formula, \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we can solve for \( t \). This step is critical, as knowing the flight time allows us to compute the horizontal distance the projectile will cover.Accurate calculations help ensure the hay bales land precisely at the location where the cattle are stranded.