Problem 93 - Precalculus with Limits

Question

A batted baseball leaves the bat at an angle of \( \theta \) with the horizontal and an initial velocity of \( v_0 = 100 \) feet per second. The ball is caught by an outfielder \( 300 \) feet from home plate (see figure).Find \( \theta \) if the range \( r \) of a projectile is given by \( r = \dfrac{1}{32} v_0^2 \sin 2 \theta \).

Step-by-Step Solution

Answer

\(\theta = \dfrac{1}{2}\sin^{-1}(0.96)\)

Step 1: Insert Given Values Into The Equation

Substitute the given values into the equation for the range of a projectile to get the sine of twice the angle. So, \(300 = \dfrac{1}{32} \times (100)^2 \times \sin 2\theta\).

Step 2: Simplification Of The Equation

Simplify the equation to solve for \( \sin 2\theta \), resulting in \( \sin 2\theta = \dfrac{300 \times 32}{100^2} = 0.96 \).

Step 3: Calculate The Angle

Using inverse sine function to calculate the angle. So, \( 2\theta = \sin^{-1}(0.96)\).

Step 4: Obtain \(\theta\)

Finally, divide by 2 to get \(\theta\). So, \( \theta = \dfrac{1}{2}\sin^{-1}(0.96)\).

Key Concepts

Range of a Projectile

Projectile motion is a fascinating area of physics that involves objects being thrown or projected. One of the critical aspects to understand is the range of a projectile. The range refers to how far a projectile will travel horizontally before coming back down to the ground.
The range can be calculated using the formula:\[ r = \dfrac{1}{32} v_0^2 \sin 2\theta \]
Here:

• \( r \) is the range.
• \( v_0 \) is the initial velocity with which the projectile is launched (in this problem, 100 ft/s).
• \( \theta \) is the angle of launch with respect to the horizontal.
• \( \sin 2\theta \) represents the sine of twice the angle of launch, which influences the parabolic path.

A higher initial velocity or a more optimized launch angle can potentially increase the range. Solving for the angle is a matter we will explore using trigonometric functions.

Trigonometric Functions

Trigonometric functions are mathematical functions related to the angles of a right triangle. In projectile motion, these functions help relate angles to directional vectors, such as velocity.
For example, the sine function gives the ratio of the length of the side of the triangle opposite the angle to the hypotenuse. Here, it is crucial for determining the range:\( \sin 2\theta \).
Some key trigonometric functions include:

• **Sine (\sin)**: Opposite side over hypotenuse.
• **Cosine (\cos)**: Adjacent side over hypotenuse.
• **Tangent (\tan)**: Opposite side over adjacent side.

In our problem, we simplify the equation by calculating \( \sin 2\theta\) and find its value as 0.96. This result is crucial for exploring the angle \( \theta \) using inverse trigonometric functions.

Inverse Sine Function

The inverse sine function, often written as \( \sin^{-1} \) or \( \arcsin \), is used to find angles when the sine of that angle is known. Inverse functions "undo" what the original function does, making them incredibly useful in solving equations involving trigonometric functions.
In our problem:

• We know that \( \sin 2\theta = 0.96 \).
• To find \( 2\theta \), we compute \( \sin^{-1}(0.96) \).

This step gives us twice the angle, \( 2\theta \). By dividing this result by 2, we derive \( \theta \), the angle with which the baseball was hit.
Understanding how to use the inverse sine function is vital for converting trigonometric values back to angles, especially in geometry and physics contexts.