STEM X (Beta)
Temasek Polytechnic STEMX
Table of Contents
- Projectile Motion (S3)
- Introduction
- Throw the Ball!
- Gardening Paradise
- Projectile Motion (S4)
- Introduction
- Fire at the Enemy!
- Cannonball
Introduction
Have you ever watched a basketball arcing through the air into the hoop, or noticed how a fountain's water makes a curved path as it falls back to the ground? What you're observing is a beautiful example of [b]projectile motion[/b] – a fundamental concept that connects mathematics, physics, and engineering.[br][br]Projectile motion describes the movement of an object thrown or projected into the air, subject to only the force of gravity. This motion is predictable and follows a specific path called a trajectory. The trajectory of a projectile is a parabola – a shape you might recognize from your math classes.[br][br]Let's throw an object with initial velocity [math]v\ \text{m}/\text{s}[/math], initial height [math]y_0\ \text{m}[/math], and angle of elevation [math]\theta[/math].
The height [math]y[/math] metres of the ball whose horizontal distance is [math]x[/math] metres away is given by the equation[br][br][center][math]y=ax^2+bx+c.[/math][/center]where [math]a=-\frac{g}{2v^2\left(\cos\theta\right)^2}[/math], [math]b=\tan\theta[/math], and [math]c=y_0[/math].[br][br]For this worksheet, we will fix [math]\theta=45^{\circ}[/math].
For [math]\theta=45^{\circ}[/math], what is [math]2\left(\cos\theta\right)^2[/math]?
For [math]\theta=45^{\circ}[/math], what is [math]\tan\theta[/math]?
We will therefore use the simplified equation to model the projectile motion:[br][br][center][math]y=-\frac{g}{v^2}x^2+x+y_0,[/math][/center]where [math]v\ \text{m}/\text{s}[/math] denote the initial speed, [math]y_0\ \text{m}[/math] denotes the initial height, and [math]g \approx 9.81\ \text{m}/\text{s}^2[/math] denotes the [b]gravitational constant[/b] on earth.
Introduction
Have you ever watched a basketball arcing through the air into the hoop, or noticed how a fountain's water makes a curved path as it falls back to the ground? What you're observing is a beautiful example of [b]projectile motion[/b] - a fundamental concept that connects mathematics, physics, and engineering.[br][br]Projectile motion describes the movement of an object thrown or projected into the air, subject to external forces (e.g.). This motion is predictable and follows a specific path called a trajectory. The trajectory of a simple projectile is a parabola – a shape you might recognize from your math classes.[br][br]Let's throw an object with initial velocity [math]v_0\ \mathrm{m}/\mathrm{s}[/math] initial height [math]y_0\ \text{m}[/math], and angle of elevation [math]\theta[/math].
The height [math]y[/math] metres of the ball whose horizontal distance is [math]x[/math] metres away is given by the equation[br][br][center][math]y=ax^2+bx+c.[/math][br][/center]where [math]a=-\frac{g}{2v_0^2\left(\cos\theta\right)^2}[/math], [math]b=\tan\theta[/math], and [math]c=y_0[/math].
For this worksheet, we will fix [math]v_0 = 10\ \mathrm{m}/\mathrm{s}[/math]. We will therefore use the simplified equation to model the projectile motion:[center][math]y=-\frac{g}{200\cos^2\left(\theta\right)}x^2+\left(\tan\theta\right)x+y_0,[/math][br][/center]where [math]\theta[/math] denotes the angle of elevation in [b]degrees[/b], [math]y_0\ \text{m}[/math] denotes the initial height, and [math]g \approx 9.81\ \text{m}/\text{s}^2[/math] denotes the [b]gravitational constant[/b] on Earth.