Problem 67 - Precalculus Mathematics for Calculus

Question

The range \(R\) and height \(H\) of a shot put thrown with an initial velocity of \(v_{0}\) ft's at an angle \(\theta\) are given by $$\begin{array}{l} R=\frac{v_{0}^{2} \sin (2 \theta)}{g} \\ H=\frac{v_{0}^{2} \sin ^{2} \theta}{2 g} \end{array}$$ On the earth \(q=32 \mathrm{ft} / \mathrm{s}^{2}\) and on the moon \(g=5.2 \mathrm{ft} / \mathrm{s}^{2} .\) Find the range and height of a shot put thrown under the given conditions. (a) On the earth with \(v_{0}=12 \mathrm{ft} / \mathrm{s}\) and \(\theta=\pi / 6\) (b) On the moon with \(u_{b}=12\) ft \(/ s\) and \(\theta=\pi / 6\)

Step-by-Step Solution

Answer

On Earth, \( R = \frac{9\sqrt{3}}{4} \) ft and \( H = \frac{9}{16} \) ft. On the Moon, \( R \approx 23.98 \) ft and \( H \approx 3.46 \) ft.

Step 1: Calculate the Range on Earth

Given that the initial velocity, \(v_0 \), is 12 ft/s, the angle \( \theta \) is \( \frac{\pi}{6} \), and the acceleration due to gravity, \( g \), is 32 ft/s² on Earth. We will use the formula for range, \( R = \frac{v_0^2 \sin(2\theta)}{g} \).First, calculate \( 2\theta \):\[2\theta = 2 \times \frac{\pi}{6} = \frac{\pi}{3}.\]Then, calculate \( \sin(2\theta) \):\[\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}.\]Substitute the known values into the range formula:\[R = \frac{12^2 \times \frac{\sqrt{3}}{2}}{32} = \frac{144 \times \frac{\sqrt{3}}{2}}{32} = \frac{72\sqrt{3}}{32} = \frac{9\sqrt{3}}{4} \text{ ft}.\]

Step 2: Calculate the Height on Earth

Use the formula for height, \( H = \frac{v_0^2 \sin^2\theta}{2g} \), with \( v_0 = 12 \text{ ft/s}, \theta = \frac{\pi}{6} \), and \( g = 32 \text{ ft/s}^2 \).Calculate \( \sin(\theta) \):\[\sin(\frac{\pi}{6}) = \frac{1}{2}.\]So, \( \sin^2(\theta) = \left( \frac{1}{2} \right)^2 = \frac{1}{4} \).Substitute into the formula for height:\[H = \frac{12^2 \times \frac{1}{4}}{64} = \frac{144 \times \frac{1}{4}}{64} = \frac{36}{64} = \frac{9}{16} \text{ ft}.\]

Step 3: Calculate the Range on the Moon

On the moon, the acceleration due to gravity, \( g \), is 5.2 ft/s². Use the same formula for range:\[R = \frac{12^2 \times \frac{\sqrt{3}}{2}}{5.2} = \frac{144 \times \frac{\sqrt{3}}{2}}{5.2} = \frac{72\sqrt{3}}{5.2} \text{ ft}.\]Approximate \( \sqrt{3} \approx 1.732 \) and calculate:\[R = \frac{72 \times 1.732}{5.2} \approx 23.98 \text{ ft}.\]

Step 4: Calculate the Height on the Moon

For the moon, using the height formula again:\[H = \frac{12^2 \times \frac{1}{4}}{10.4} = \frac{144 \times \frac{1}{4}}{10.4} = \frac{36}{10.4} \text{ ft}.\]Simplify to find the height:\[H = \frac{36}{10.4} \approx 3.46 \text{ ft}.\]

Key Concepts

Initial Velocity

The concept of initial velocity is fundamental in understanding projectile motion. In our problem of calculating the range and height of a shot put, the initial velocity (\(v_0\)) is the speed at which the projectile is launched.

• The initial velocity you encounter here is 12 ft/s.
• This velocity will significantly affect how far and high your shot put will go.

An understanding of initial velocity is essential because it directly influences both the horizontal and vertical components of projectile motion.
These components determine how long the projectile stays in the air and where it lands, given constant conditions such as air resistance being negligible.It is crucial to remember that without an appropriate initial velocity, the projectile might not reach the desired distance or height.

Angle of Projection

The angle of projection refers to the angle at which an object is launched into the air. It's denoted by \(\theta\) and it plays a vital role in the flight path of the projectile. In our example, \(\theta = \frac{\pi}{6}\).

• The angle determines how much of the initial velocity is distributed into horizontal and vertical motion.
• A steeper angle focuses more energy upward, increasing maximum height.
• A shallower angle enhances horizontal travel but might reduce vertical reach.

Selecting the correct angle of projection is crucial for optimizing range and height.
In this exercise, understanding how the angle affects both range and height calculations helps in fine-tuning your expectations for the projectile's trajectory.

Acceleration Due to Gravity

Gravity is the force that pulls the projectile back toward the Earth (or Moon, in this exercise). It's represented by the symbol \(g\), and it varies depending on the celestial body.

• On Earth, \(g\) is 32 ft/s².
• On the Moon, it is weaker at 5.2 ft/s².

This acceleration affects how fast a projectile falls back down, significantly impacting range and height factors.
A strong gravitational pull (like on Earth) means the projectile will not travel as far or high compared to a place with weaker gravity (like the Moon).
Understanding gravity's role in projectile motion is key to predicting the movement and adjusting either the initial velocity or angle to achieve better distance or elevation.

Trigonometric Functions

Trigonometry plays a crucial role in analyzing projectile motion. Specifically, functions like sine (\(\sin\)) are used for calculating the range and height of a projectile.

• In our formulas, \(\sin(2\theta)\) and \(\sin(\theta)\) are essential for deriving key components of motion.
• They help separate the initial velocity into horizontal and vertical components.
• Frequently, you will encounter angles in radians, as with \(\frac{\pi}{6}\) here.

Using trigonometric functions, you can compute the necessary proportions of motion that determine how far and high the projectile will travel.
Mastering these calculations ensures that you can precisely control different aspects of a projectile's path for a successful launch.