Problem 65 - Algebra and Trigonometry

Question

Throwing a Shot Put The range \(R\) and height \(H\) of a shot put thrown with an initial velocity of \(v_{0}\) ft \(t / s\) at an angle \(\theta\) are given by $$R=\frac{v_{0}^{2} \sin (2 \theta)}{g}$$ $$H=\frac{v_{0}^{2} \sin ^{2} \theta}{2 g}$$ On the earth \(g=32 \mathrm{ft} / \mathrm{s}^{2}\) 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 \(v_{0}=12 \mathrm{ft} / \mathrm{s}\) and \(\theta=\pi / 6\)

Step-by-Step Solution

Answer

On Earth: Range \(= \frac{9\sqrt{3}}{4}\) ft, Height \(= \frac{9}{16}\) ft. On Moon: Range \(\approx 24.72\) ft, Height \(\approx 3.46\) ft.

Step 1: Identify Given Values on Earth

For the earth, we have the following given values: \(v_0 = 12\) ft/s, \(\theta = \frac{\pi}{6}\), and \(g = 32\) ft/s². These values will be used to calculate the range and height on Earth.

Step 2: Calculate the Range on Earth

Using the range formula \(R = \frac{v_0^2 \sin(2\theta)}{g}\), substitute the values: \[R = \frac{12^2 \sin(2 \cdot \frac{\pi}{6})}{32} = \frac{144 \sin(\frac{\pi}{3})}{32}.\] Since \(\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}\), the equation becomes \[R = \frac{144 \cdot \frac{\sqrt{3}}{2}}{32} = \frac{72\sqrt{3}}{32} = \frac{9\sqrt{3}}{4} \text{ ft}.\]

Step 3: Calculate the Height on Earth

For the height, use the formula \(H = \frac{v_0^2 \sin^2 \theta}{2g}\). Substitute the values: \[H = \frac{12^2 \sin^2\left(\frac{\pi}{6}\right)}{2 \times 32} = \frac{144 \cdot \left(\frac{1}{2}\right)^2}{64} = \frac{144 \cdot \frac{1}{4}}{64}.\] Simplifying gives \[H = \frac{36}{64} = \frac{9}{16} \text{ ft}.\]

Step 4: Identify Given Values on the Moon

For the moon, the given values are still \(v_0 = 12\) ft/s and \(\theta = \frac{\pi}{6}\), but the gravitational acceleration changes to \(g = 5.2\) ft/s².

Step 5: Calculate the Range on the Moon

Using the same range formula with the moon's gravity: \[R = \frac{12^2 \sin(2 \cdot \frac{\pi}{6})}{5.2} = \frac{144 \cdot \frac{\sqrt{3}}{2}}{5.2}.\] Simplifying gives \[R = \frac{72\sqrt{3}}{5.2} \approx 24.72 \text{ ft}.\]

Step 6: Calculate the Height on the Moon

Substitute into the height formula with moon gravity: \[H = \frac{12^2 \sin^2\left(\frac{\pi}{6}\right)}{2 \times 5.2} = \frac{144 \cdot \frac{1}{4}}{10.4}.\] Simplifying gives \[H = \frac{36}{10.4} \approx 3.46 \text{ ft}.\]

Key Concepts

Range Formula

The range formula is essential in determining the horizontal distance a projectile travels. In the case of a shot put, the formula can be expressed as \(R = \frac{v_{0}^{2} \sin(2\theta)}{g}\). Here, \(v_0\) represents the initial velocity of the shot put, \(\theta\) is the angle of projection relative to the horizontal, and \(g\) is the acceleration due to gravity.

The range formula is derived from the physics of projectile motion, assuming there is no air resistance.

• The double angle in \(\sin(2\theta)\) specifically relates to the horizontal component of the motion.
• The initial speed \(v_0\) influences how far the projectile can go.
• As gravitational acceleration \(g\) increases, the range decreases because the projectile comes back to the ground quicker.

For students, understanding this formula helps in analyzing problems where the horizontal distance needs to be calculated, especially under varying gravitational conditions like on Earth versus the Moon.

Height Formula

The height formula in projectile motion is used to find how high a projectile, such as a shot put, will rise at its peak. The equation is given by \(H = \frac{v_0^2 \sin^2 \theta}{2g}\). This formula considers several key factors:

• \(\sin^2 \theta\) represents the vertical component of the launch angle concentrated on height.
• \(v_0\) is the initial speed, where a higher speed results in a higher trajectory.
• The division by \(2g\) accounts for the deceleration due to gravity, halving the effect because gravity pulls the shot back down.

When employing this formula, one's understanding should expand to recognizing how changes in gravity, for instance from Earth to the Moon, impact the height, as less gravitational force permits the shot put to reach a higher altitude.

Gravitational Acceleration

Gravitational acceleration is a fundamental concept in physics that significantly affects projectile motion. On Earth, this is approximately \(32\, \text{ft/s}^2\), whereas on the Moon it drops to about \(5.2\, \text{ft/s}^2\). This value determines how quickly a projectile comes back towards the ground once it is propelled upwards.
Consider these points:

• A higher gravitational acceleration results in a shorter time spent in the air and therefore a shorter range and maximum height.
• In contrast, a lower gravitational force, like on the Moon, allows the projectile to travel further and higher because less force pulls it back down.
• Understanding this helps in solving problems involving varying gravitational scenarios.

By comprehending gravitational acceleration, students can relate the impact of gravity on both the range and height of any projectile.

Trigonometric Functions

Trigonometric functions play a crucial role in analyzing projectile motion because they help break down the velocity and angles into components that describe motion. In the range and height formulas for projectile motion, sine functions \(\sin(2\theta)\) and \(\sin^2\theta\) appear frequently.

• \(\sin(2\theta)\) helps us understand the impact of the angle on horizontal motion, often maximizing range at \(45^\circ\) due to the properties of the sine function.
• \(\sin^2\theta\) comes in handy to concentrate on vertical displacement, influencing how high the projectile will go.
• The sine function, especially in physics, is crucial when vectors are involved.

When looked at collectively, these trigonometric components facilitate deeper insight into how effectively a projectile has been launched, aimed, and how it interacts with surrounding forces.