Problem 67 - Algebra and Trigonometry

Question

Throwing a Shot Put The range \(R\) and height \(H\) of a shot put thrown with an initial velocity of \(v_{0} \mathrm{ft} / \mathrm{s}\) 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 \(g=32 \mathrm{ft} / \mathrm{s}^{2}\) 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: \( R \approx 3.897 \text{ ft}, H = 0.5625 \text{ ft} \). On Moon: \( R \approx 24.728 \text{ ft}, H = 3.462 \text{ ft} \).

Step 1: Identify the known values for Earth

Given that the shot put is thrown on Earth, use the values: initial velocity \( v_0 = 12 \text{ ft/s}\), angle \( \theta = \frac{\pi}{6} \), and acceleration due to gravity \( g = 32 \text{ ft/s}^2 \).

Step 2: Calculate the Range on Earth

Use the formula for range: \( R = \frac{v_0^2 \sin (2 \theta)}{g} \). Substitute the known values: \[R = \frac{12^2 \sin(2 \times \frac{\pi}{6})}{32}\]Since \( \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2} \), the calculation becomes:\[R = \frac{144 \times \frac{\sqrt{3}}{2}}{32} = \frac{72\sqrt{3}}{32} = \frac{9\sqrt{3}}{4} \approx 3.897 \text{ ft}\]

Step 3: Calculate the Height on Earth

Use the formula for height: \( H = \frac{v_0^2 \sin^2 \theta}{2g} \). Substitute the known values:\[H = \frac{12^2 \sin^2(\frac{\pi}{6})}{2 \times 32}\]Since \( \sin(\frac{\pi}{6}) = \frac{1}{2} \), the calculation becomes:\[H = \frac{144 \times \left(\frac{1}{2}\right)^2}{64} = \frac{36}{64} = \frac{9}{16} = 0.5625 \text{ ft}\]

Step 4: Identify the known values for Moon

Now consider the shot put on the Moon using the values: initial velocity \( v_0 = 12 \text{ ft/s}\), angle \( \theta = \frac{\pi}{6} \), and acceleration due to gravity \( g = 5.2 \text{ ft/s}^2 \).

Step 5: Calculate the Range on Moon

Use the same range formula: \( R = \frac{v_0^2 \sin (2 \theta)}{g} \). Substitute the values for the Moon:\[R = \frac{12^2 \sin(2 \times \frac{\pi}{6})}{5.2}\]With the same trigonometric result:\[R = \frac{144 \times \frac{\sqrt{3}}{2}}{5.2} = \frac{72\sqrt{3}}{5.2} \approx 24.728 \text{ ft}\]

Step 6: Calculate the Height on Moon

Use the height formula: \( H = \frac{v_0^2 \sin^2 \theta}{2g} \). Substitute the known values:\[H = \frac{12^2 \sin^2(\frac{\pi}{6})}{2 \times 5.2}\]Continue evaluating as before:\[H = \frac{144 \times \left(\frac{1}{2}\right)^2}{10.4} = \frac{36}{10.4} = 3.462 \text{ ft}\]

Key Concepts

Range formula

The range formula is a crucial aspect of projectile motion, determining how far an object will travel horizontally before it hits the ground. The formula to calculate the range \( R \) is given by:\[ R = \frac{v_0^2 \sin(2 \theta)}{g} \]where:- \( v_0 \) is the initial velocity of the object,- \( \theta \) is the angle of projection,- \( g \) is the gravitational acceleration.The formula considers both the speed and angle at which the shot put is launched as well as the force of gravity.

• On Earth, with standard gravitational acceleration \( g = 32 \text{ ft/s}^2 \), the range depends heavily on the angle. Optimum range is typically achieved at a 45-degree angle.
• On the Moon, where \( g = 5.2 \text{ ft/s}^2 \), the same principles apply, but the reduced gravitational force allows objects to travel further.

By understanding the range formula, one can predict and optimize the distance a projectile will travel.

Angle of projection

The angle of projection is one of the most influential factors in determining the path of a projectile. In the range formula \( R = \frac{v_0^2 \sin(2 \theta)}{g} \), the term \( \sin(2 \theta) \) indicates how the angle affects the range.

• A 45-degree angle is often considered ideal for maximizing range under uniform conditions.
• Angles less than 45 degrees tend to increase vertical height but reduce range.
• Conversely, angles greater than 45 degrees lead to greater height but a shorter horizontal distance, making it less efficient for range.

Understanding the angle of projection helps in making strategic decisions about the shot put's initial launch. By adjusting \( \theta \), throwers can fine-tune their performance to suit different conditions and gravitational influences like those on the Moon or Earth.

Gravitational acceleration

Gravitational acceleration \( g \) is a key factor in projectile motion, representing the rate at which objects accelerate towards the center of a celestial body.On Earth, \( g \) is approximately \( 32 \text{ ft/s}^2 \), affecting how quickly a projectile returns to ground after being thrown.

• On the Moon, the gravitational pull is much weaker, with \( g = 5.2 \text{ ft/s}^2 \). This lower gravity allows objects to stay airborne longer and travel further, provided they have the same initial velocity as on Earth.

Gravitational acceleration directly influences both the maximum range and height a projectile can achieve. In calculations for projectile motion, \( g \) is pivotal in determining how trajectories differ across various environments. Whether on the Moon or Earth, understanding \( g \) helps in analyzing and predicting the behavior of projectiles under different gravitational forces.