Problem 56 - Precalculus with Calculus Previews

Question

Putting the Shot The range of a shot put released from a height \(h\) above the ground with an initial velocity \(v_{0}\) at an angle \(\theta\) to the horizontal can be approximated by: $$ R=\frac{v_{0} \cos \theta}{g}\left(v_{0} \sin \theta+\sqrt{v_{0}^{2} \sin ^{2} \theta+2 g h}\right) $$ where \(g\) is the acceleration due to gravity. If \(v_{\mathrm{o}}=\) \(13.7 \mathrm{~m} / \mathrm{s}, \theta=40^{\circ},\) and \(g=9.81 \mathrm{~m} / \mathrm{s}^{2},\) compare the ranges achieved for the release heights (a) \(h=2.0\) \(\mathrm{m}\) and (b) \(h=2.4 \mathrm{~m} .\) (c) Explain why an increase in \(h\) yields an increase in \(R\) if the other parameters are held fixed. (d) What does this imply about the advantage that height gives a shot-putter?

Step-by-Step Solution

Answer

(a) 20.27 m; (b) 20.81 m; Increased height allows a longer trajectory.

Step 1: Substitute Known Values for Case (a)

Plug in the values for \( v_0 = 13.7 \, \text{m/s}\), \( \theta = 40^\circ \), \( h = 2.0 \, \text{m} \), and \( g = 9.81 \, \text{m/s}^2 \) into the range formula:\[ R = \frac{v_0 \cos \theta}{g} \left( v_0 \sin \theta + \sqrt{v_0^2 \sin^2 \theta + 2gh} \right) \]Calculate \( \cos \theta \) and \( \sin \theta \):- \( \cos 40^\circ = 0.766 \)- \( \sin 40^\circ = 0.643 \)Substitute these values into the equation to find the range.

Step 2: Calculate the Terms for Case (a)

Calculate each part of the expression. Start by finding components:\( v_0 \cos \theta = 13.7 \times 0.766 = 10.4958 \, \text{m/s} \)\( v_0 \sin \theta = 13.7 \times 0.643 = 8.7991 \, \text{m/s} \)Next, compute the square root term:\[ \sqrt{v_0^2 \sin^2 \theta + 2gh} = \sqrt{(13.7)^2 \times (0.643)^2 + 2 \times 9.81 \times 2.0} \approx 10.412 \, \text{m/s} \]Now, substitute back into the range formula to get the range for \( h = 2.0 \, \text{m} \):\[ R = \frac{10.4958}{9.81} \left( 8.7991 + 10.412 \right) \approx 20.27 \, \text{m} \]

Step 3: Substitute Known Values for Case (b)

Now substitute the values for \( h = 2.4 \, \text{m} \) keeping all other parameters the same. Reuse \( v_0 \cos \theta \) and \( v_0 \sin \theta \) from Step 2:Calculate the new square root term:\[ \sqrt{v_0^2 \sin^2 \theta + 2gh} = \sqrt{(13.7)^2 \times (0.643)^2 + 2 \times 9.81 \times 2.4} \approx 10.629 \, \text{m/s} \]

Step 4: Calculate the Terms for Case (b)

Substitute into the range formula:\[ R = \frac{10.4958}{9.81} \left( 8.7991 + 10.629 \right) \]Once all values are substituted and calculations are completed:\[ R \approx 20.81 \, \text{m} \]

Step 5: Comparison of Ranges

So, the range for \( h = 2.0 \, \text{m} \) is approximately \( 20.27 \, \text{m} \) while the range for \( h = 2.4 \, \text{m} \) is approximately \( 20.81 \, \text{m} \). Clearly, increasing the height \( h \) results in a longer range.

Step 6: Explanation of Height Effect on Range

Increasing the height \( h \) increases the square root term \( \sqrt{v_0^2 \sin^2 \theta + 2gh} \) in the range formula, which in turn increases the entire range \( R \). This occurs because additional height allows the shot to travel a longer vertical trajectory before hitting the ground, allowing more time for horizontal travel at the same velocity.

Step 7: Implication for Shot-Putters

A greater release height gives a shot-putter an advantage because it increases the range. Therefore, taller athletes or those who can effectively utilize higher release points can potentially achieve greater shot put distances.

Key Concepts

Range Formula

The range formula is a crucial tool for calculating the distance an object will travel in projectile motion. In the context of shot put, it tells us how far the shot will travel if released with a specific initial velocity and angle. The formula is given by:\[ R = \frac{v_0 \cos \theta}{g} \left( v_0 \sin \theta + \sqrt{v_0^2 \sin^2 \theta + 2gh} \right) \]Here's what each part represents:

• \( v_0 \) is the initial velocity of the shot.
• \( \theta \) is the angle of release.
• \( g \) stands for acceleration due to gravity, making downward motion inevitable.
• \( h \) is the release height which affects the time the shot spends in the air.

By solving this equation, we obtain the range \( R \) which is how far horizontally the shot goes before touching the ground. A higher range implies a longer distance for the shot put, which is essential for achieving wins in competitions.

Shot Put

Shot put is a track and field event involving 'putting' a heavy spherical object, called a shot, as far as possible. Unlike throwing, the shot put is a unique technique aiming to maximize the range of the projectile.
Several factors contribute to success in shot put, including:

• The initial speed at which the shot is released.
• The angle of release, which ideally should be around 40-45 degrees for optimal performance.
• The height from which the shot is released. This aspect, often overlooked, can significantly impact the distance achieved.

By mastering these elements, athletes can optimize their throws to gain an upper hand in competitive settings. Therefore, understanding projectile motion and its intricacies such as angle and initial velocity, is critical for those pursuing excellence in shot put.

Release Height Effect

Release height has a significant impact on projectile range in events like the shot put. When a shot is released from a higher point, the formula shows it has more time to travel vertically and thus cover more horizontal distance before landing. This happens due to:

• An increase in height \( h \) enhances the term \( \sqrt{v_0^2 \sin^2 \theta + 2gh} \), providing a larger range.
• The increased vertical component allows the shot to stay airborne longer, extending the time it can travel horizontally.

This effect explains why taller athletes or those who can create higher launch points often have a competitive advantage. It is not just brute strength but the strategic use of physics and height that leads to successful shot puts. This principle underscores the importance of technique and body mechanics in athletics.