Problem 123 - Algebra and Trigonometry Real Mathematics, Real People

Question

The range of a projectile fired at an angle \(\theta\) with the horizontal and with an initial velocity of \(v_{0}\) feet per second is given by $$r=\frac{1}{32} v_{0}^{2} \sin 2 \theta$$ where \(r\) is measured in feet. An athlete throws a javelin at 75 feet per second. At what angle must the athlete throw the javelin so that the javelin travels 130 feet?

Step-by-Step Solution

Answer

The athlete must throw the javelin at an angle of \(\frac{1}{2} \sin^{-1} \left(\frac{130}{\frac{1}{32} \cdot 75^2}\right)\) degrees from the horizontal direction to achieve a range of 130 feet.

Step 1: Insert the given values into the specified formula

Rewrite the formula \(r=\frac{1}{32} v_{0}^{2} \sin 2 \theta\) by substituting \(v_0 = 75 \, \text{feet per second}\) and \(r = 130 \, \text{feet}\): \[130 = \frac{1}{32} \cdot (75^2) \cdot \sin 2 \theta\]

Step 2: Solve for the \(\sin 2 \theta\)

Isolate \(\sin 2 \theta\) by dividing both sides of the equation by \( \frac{1}{32} \cdot 75^2\): \[\sin 2 \theta = \frac{130}{\frac{1}{32} \cdot 75^2}\]

Step 3: Use the Inverse Sine Function

Since we now solved for \(\sin 2 \theta\), we must apply the inverse sine function \(\sin^{-1}\) to find \(\theta\): \[2 \theta = \sin^{-1} \left( \frac{130}{\frac{1}{32} \cdot 75^2} \right)\]

Step 4: Solve for \(\theta\)

Finally divide 2 from both sides of the equation to find \(\theta\): \[\theta = \frac{1}{2} \times \sin^{-1} \left( \frac{130}{\frac{1}{32} \cdot 75^2} \right)\]

Key Concepts

Angle of Projection

The angle of projection is crucial in determining how far a projectile will travel. It refers to the angle at which an object is launched relative to the horizontal. Choosing the right angle can significantly affect the range of the projectile. The angle helps to balance vertical and horizontal motion, optimizing how far and how high the projectile will go.

Generally, a 45-degree angle is considered optimal for achieving maximum range in a vacuum, offering a perfect balance between height and distance. However, in real-world scenarios, other factors like air resistance can affect this ideal projection angle.

Initial Velocity

Initial velocity is the speed at which a projectile is launched. This speed determines how far and how high the object will go. Initial velocity has two components:

• Horizontal velocity—affects how far the object travels.
• Vertical velocity—affects how high the object rises.

The initial velocity, denoted as \(v_0\), plays a crucial role as it directly influences the range of the projectile. In the given problem, the initial velocity is 75 feet per second, which is vital for calculating the exact angle needed to reach the desired range.

Inverse Sine Function

The inverse sine function, also known as arcsin, is used to determine the angle whose sine value is given. In projectile motion problems, after isolating \(\sin 2\theta\) in the equation, we use the inverse sine function to find the angle \(\theta\).

Mathematically, this is expressed as \(\theta = \frac{1}{2} \times \sin^{-1}(\sin 2\theta)\). This function is critical for solving angle-related problems in trigonometry and plays a pivotal role in determining the precise angle for projectile range.

Range of Projectile

The range of a projectile is the horizontal distance it covers during its flight. It is determined by the initial velocity, the angle of projection, and gravity. In a vacuum, the range \(r\) can be calculated using the formula: \[r = \frac{1}{32} v_{0}^{2} \sin 2 \theta\] This equation shows how each factor impacts the distance:

• \(v_0^2\)—represents the influence of initial speed squared.
• \(\sin 2 \theta\)—captures the effect of the angle on the range.

Understanding these components helps in predicting and optimizing the path of the projectile for a specific target distance.