Problem 5 - Thomas Calculus

Question

Shot put An athlete puts a \(16-16\) shot at an angle of \(45^{\circ}\) to the horizontal from 6.5 ft above the ground at an initial speed of 44 \(\mathrm{ft} / \mathrm{sec}\) as suggested in the accompanying figure. How long after launch and how far from the inner edge of the stopboard does the shot land? Graph cannot copy

Step-by-Step Solution

Answer

The shot lands 66.27 ft from the inner edge of the stopboard after 2.13 seconds.

Step 1: Breakdown of Motion Components

The motion of the shot put can be broken into two components: horizontal motion and vertical motion. We'll use the initial velocity to find these components: - Horizontal velocity: \( v_x = 44 \cos(45^{\circ}) \).- Vertical velocity: \( v_y = 44 \sin(45^{\circ}) \).Given the angle, both components equal \( \frac{44}{\sqrt{2}} \approx 31.11 \text{ ft/sec} \).

Step 2: Calculate Time of Flight

To find how long the shot is in the air, we use the vertical motion equation considering initial height. The equation is:\[ s = v_{y}t - \frac{1}{2}gt^2 + h_0 \]Setting \(s = 0\) for when it hits the ground, \(h_0 = 6.5 \text{ ft}\), and \(g = 32 \text{ ft/sec}^2\), we solve:\[ 0 = 31.11t - 16t^2 + 6.5 \]Solving this quadratic equation gives us the time of flight \( t \approx 2.13 \text{ sec} \).

Step 3: Calculate Horizontal Distance Traveled

Using the time of flight, we can calculate the horizontal distance:\[ d = v_x t = 31.11 \times 2.13 = 66.27 \text{ ft} \].

Key Concepts

Kinematics

Understanding kinematics is essential for analyzing motion, like the shot put exercise. Kinematics is a branch of physics that describes the motion of points, objects, and systems without considering the forces that cause them.
In projectile motion, like in the shot put, kinematic equations allow us to determine variables such as time, speed, and position.
For projectiles, the key assumptions are that motion occurs in a two-dimensional plane and gravity is the only force acting on the object.

• Displacement, velocity, and acceleration are the primary descriptors of motion.
• Kinematics equations simplify motion into two separate components, horizontal and vertical, which helps in accurately solving projectile motion problems.

By applying these principles, we can effectively analyze the shot's trajectory and determine how long it stays in the air and the distance it covers before hitting the ground.

Quadratic Equation

Quadratic equations are pivotal for calculating the vertical motion of projectiles. In our exercise, the equation determining how long the shot is in the air is quadratic.
This type of equation generally looks like this: \[ ax^2 + bx + c = 0 \]In the context of projectile motion:

• \( a \) is often related to half of the acceleration due to gravity.
• \( b \) is the initial velocity component vertically.
• \( c \) is the initial height from which the object is projected.

The solution to this quadratic equation provides the time of flight. In the shot put exercise, we use the equation \[ 0 = 31.11t - 16t^2 + 6.5 \]This helps calculate the time until the shot hits the ground, which is approximately 2.13 seconds. Understanding how to solve quadratic equations is essential for finding the time duration of the shot's motion.

Horizontal and Vertical Components

Breaking motion into horizontal and vertical components is a key strategy in solving projectile motion problems. Each component acts independently of the other.
For the shot put example:

• The horizontal component (\( v_x \)) is determined using the cosine of the launch angle:

\[ v_x = 44 \cos(45^{\circ}) \approx 31.11 \text{ ft/sec} \]

• The vertical component (\( v_y \)) is determined using the sine:

\[ v_y = 44 \sin(45^{\circ}) \approx 31.11 \text{ ft/sec} \]The independence of these components allows one to focus separately: horizontal motion is constant since it is independent of gravity, while vertical motion is influenced by gravity. This independence aids in resolving how far and how long a projectile travels.

Initial Velocity

Initial velocity plays a crucial role in determining the trajectory and overall behavior of a projectile. It's the velocity the object possesses at the moment of launch.
With a given magnitude and direction, initial velocity affects how far and high the projectile will travel. In the shot put exercise, our initial velocity is 44 ft/sec at a 45-degree angle to the horizontal.

• This results in equal horizontal and vertical components because \( \cos(45^{\circ}) \) and \( \sin(45^{\circ}) \) are equal.

Knowing initial velocity helps in calculating the required time in the air and how far the projectile will land.
It serves as the initial input when applying kinematic equations to break motion into horizontal and vertical components. Understanding initial velocity is foundational to analyzing and predicting projectile paths accurately.