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Chapter 4: Problem 70

A shot-putter throws the shot with an initial speed of \(12.2 \mathrm{m} / \mathrm{s}\) from a height of \(5.15 \mathrm{ft}\) above the ground. What is the range of the shot if the launch angle is (a) \(20.0^{\circ}\) (b) \(30.0^{\circ},\) or (c) \(40.0^{\circ} ?\)

Short Answer

Expert verified
Range is 13.44 m for 20掳, 14.84 m for 30掳, and 15.02 m for 40掳.

Step by step solution

01

Convert Height to Meters

The initial height of the shot is given in feet. To use consistent units, we'll convert this height from feet to meters. We know that 1 foot is approximately 0.3048 meters, so we multiply:\[ 5.15 \text{ ft} \times 0.3048 \frac{\text{m}}{\text{ft}} \approx 1.57 \text{ m} \].
02

Understand Projectile Motion

The shot follows a projectile motion. The key equations for projectile motion are:1. Horizontal range \( R = \frac{v_0^2 \sin(2\theta)}{g} \) if launched and landed at the same height.2. For a launch from a height, we adjust using \[ R = \frac{v_0 \cos(\theta)}{g} \left(v_0 \sin(\theta) + \sqrt{(v_0 \sin(\theta))^2 + 2gy}\right) \].
03

Use Values for Launch Angle (a) 20掳

Use the given speed \( v_0 = 12.2 \text{ m/s} \), launch angle \( \theta = 20^{\circ} \), and initial height \( y = 1.57 \text{ m} \). The horizontal range is given by: \[ R = \frac{12.2 \cos(20^{\circ})}{9.81} \left(12.2 \sin(20^{\circ}) + \sqrt{(12.2 \sin(20^{\circ}))^2 + 2 \times 9.81 \times 1.57}\right) \].
04

Calculate for 20掳

Calculate the individual parts: \( \sin(20^{\circ}) \approx 0.342 \) and \( \cos(20^{\circ}) \approx 0.940 \). Substitute into the equation:\[ R = \frac{12.2 \times 0.940}{9.81} \left(12.2 \times 0.342 + \sqrt{(12.2 \times 0.342)^2 + 2 \times 9.81 \times 1.57}\right) \].
05

Use Values for Launch Angle (b) 30掳

Similarly, use \( \theta = 30^{\circ} \). The range equation becomes:\[ R = \frac{12.2 \cos(30^{\circ})}{9.81} \left(12.2 \sin(30^{\circ}) + \sqrt{(12.2 \sin(30^{\circ}))^2 + 2 \times 9.81 \times 1.57}\right) \].
06

Calculate for 30掳

Calculate the individual parts: \( \sin(30^{\circ}) = 0.5 \) and \( \cos(30^{\circ}) \approx 0.866 \). Substitute into the equation:\[ R = \frac{12.2 \times 0.866}{9.81} \left(12.2 \times 0.5 + \sqrt{(12.2 \times 0.5)^2 + 2 \times 9.81 \times 1.57}\right) \].
07

Use Values for Launch Angle (c) 40掳

Now, use \( \theta = 40^{\circ} \). The range is given by:\[ R = \frac{12.2 \cos(40^{\circ})}{9.81} \left(12.2 \sin(40^{\circ}) + \sqrt{(12.2 \sin(40^{\circ}))^2 + 2 \times 9.81 \times 1.57}\right) \].
08

Calculate for 40掳

Calculate the individual components: \( \sin(40^{\circ}) \approx 0.643 \) and \( \cos(40^{\circ}) \approx 0.766 \). Substitute into the equation:\[ R = \frac{12.2 \times 0.766}{9.81} \left(12.2 \times 0.643 + \sqrt{(12.2 \times 0.643)^2 + 2 \times 9.81 \times 1.57}\right) \].
09

Results for Each Angle

After calculating for each angle, you find the range for (a) 20掳 is approximately 13.44 m, for (b) 30掳 is approximately 14.84 m, and for (c) 40掳 is approximately 15.02 m.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Angle of Launch
Understanding the angle of launch is crucial in projectile motion problems. It is the angle at which an object is projected into the air. This angle affects both the horizontal range and the maximum height that the projectile reaches.
  • A higher launch angle, such as 60掳, results in a higher but shorter path for the projectile.
  • A lower angle, like 20掳, gives a longer range but less height.
  • The optimal angle for maximum range, assuming a level launch and landing, is 45掳.
In the problem, different angles (20掳, 30掳, and 40掳) are used to see how the range changes. This helps students understand how important the launch angle is in determining where the shot lands.
Horizontal Range
The horizontal range is the distance a projectile covers along the horizontal axis from launch to landing. It largely depends on the initial velocity and the angle of launch.
  • Consider how the formula changes when launching from a height, as it affects both the time and distance.
  • The general equation for horizontal range without any height difference is given by \( R = \frac{v_0^2 \sin(2\theta)}{g} \), where \( v_0 \) is the initial velocity and \( g \) is the acceleration due to gravity.
  • When an initial height is added, adjust with extra components to account for the projectile's drop: \[ R = \frac{v_0 \cos(\theta)}{g} \left( v_0 \sin(\theta) + \sqrt{(v_0 \sin(\theta))^2 + 2gy} \right) \]
With initial heights, the equation compensates for the vertical drop, which in turn affects the travel distance. Calculating the range for each angle given in the problem shows how the trajectory changes based on angle adjustments.
Initial Velocity
Initial velocity is the speed at which the projectile is launched. In our context of projectile motion, it has both horizontal and vertical components that impact the object's trajectory.
  • The initial velocity \( v_0 \) splits into horizontal \( v_{0x} = v_0 \cos(\theta) \) and vertical \( v_{0y} = v_0 \sin(\theta) \) components.
  • The horizontal component influences how far the projectile will travel before hitting the ground.
  • The vertical component affects how high it will go and eventually the duration it stays airborne.
The given problem uses an initial speed of 12.2 m/s, affecting both the height it reaches and the distance it covers, depending on the angle of launch.
Conversion of Units
Conversion of units is an essential technique in physics problems for consistency in calculations. This problem provides initial height in feet, which needs conversion to meters to use standard equations.
  • Conversion factor: 1 foot = 0.3048 meters.
  • Given height: 5.15 feet is converted to meters as follows:\( 5.15 \text{ ft} \times 0.3048 \frac{\text{m}}{\text{ft}} = 1.57 \text{ m} \)
  • Always ensure units for velocity, distance, and acceleration (m/s虏) are consistent to avoid errors.
Consistently using metric units simplifies calculations as many physics problems, especially in projectile motion, are designed around these standard units.

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