Introduction to Gravity and Projectiles
Hi there! Welcome to one of the most exciting parts of Mechanics. Have you ever wondered how a footballer knows exactly how hard to kick a ball to reach a teammate, or how an archer aims at a target? They are all using the principles of Gravity and Projectile Motion.
In this chapter, we are going to look at how objects move when the only force acting on them is their own weight. Don't worry if Mechanics feels a bit "heavy" at first—once you see the patterns, it becomes much more manageable! We will break everything down into simple steps, from vertical drops to the curved paths of basketballs.
1. Understanding Weight and Acceleration due to Gravity
Before we start moving things, we need to understand the force pulling them down. According to the OCR syllabus, we model Gravity as a constant acceleration.
What is Weight?
Weight (\(W\)) is a force. It is the result of gravity pulling on an object's mass. We use a simple version of Newton's Second Law (\(F=ma\)) to calculate it:
\(W = mg\)
Where:
- \(m\) is the Mass of the object (in kg).
- \(g\) is the Acceleration due to Gravity.
The Value of \(g\)
In your exams, unless stated otherwise, you should use \(g = 9.8 \text{ ms}^{-2}\).
Did you know? Even though we treat \(g\) as a constant, it actually changes slightly depending on where you are on Earth! However, for our models, we assume it's always the same.
Modeling Assumptions
To make the math easier, we use the Particle Model. This means we treat objects (like a piano or a tennis ball) as a single point. This allows us to ignore:
- Air resistance (we assume the object is moving in a vacuum).
- Spin or Rotation.
- Shape or Surface Area.
Quick Review: Weight is a force (\(W=mg\)), and gravity causes a constant downward acceleration of \(9.8 \text{ ms}^{-2}\).
2. Vertical Motion Under Gravity
When an object is moving strictly up or down, we use our SUVAT equations. The most important thing here is to pick a direction to be positive and stick to it!
Using Vectors for Gravity
In two dimensions, we often write acceleration as a column vector. Since gravity only acts downwards (vertically), there is zero horizontal acceleration:
\(\mathbf{a} = \begin{pmatrix} 0 \\ -g \end{pmatrix}\) or \(\mathbf{a} = -g\mathbf{j}\)
Example: If you throw a ball straight up, its acceleration at the very top of its flight is still \(9.8 \text{ ms}^{-2}\) downwards, even though its velocity is zero for a split second!
Common Mistake to Avoid: Forgetting that \(g\) is always acting downwards. If you choose "Up" as positive, your acceleration must be \(-9.8\). If you choose "Down" as positive, your acceleration is \(+9.8\).
3. Projectile Motion: Moving in Two Dimensions
A Projectile is any object thrown or launched into the air. Its path is a curve called a parabola. The secret to solving any projectile problem is this: Horizontal and Vertical motions are completely independent.
Step-by-Step: Resolving the Initial Velocity
If an object is launched with an initial speed \(u\) at an angle \(\theta\) to the horizontal, we must split it into two components:
- Horizontal component (\(u_x\)): \(u \cos \theta\)
- Vertical component (\(u_y\)): \(u \sin \theta\)
Memory Aid: "Cos is 'cross" (Horizontal) and "Sin is 'high" (Vertical).
The Two Rules of Projectiles
Because we ignore air resistance:
1. Horizontally: There is no acceleration (\(a=0\)). The horizontal velocity never changes.
2. Vertically: There is constant acceleration (\(a = -g\)). We use SUVAT.
Horizontal Equation:
\(x = (u \cos \theta)t\)
Vertical Equations:
\(v_y = u \sin \theta - gt\)
\(y = (u \sin \theta)t - \frac{1}{2}gt^2\)
Key Takeaway: Time (\(t\)) is the "bridge" between the horizontal and vertical sides. If you find time using one, you can use it in the other!
4. Important Projectile Milestones
There are three specific things exam questions love to ask for: Greatest Height, Time of Flight, and Horizontal Range.
1. Greatest Height (\(H\))
At the very top of the curve, the object stops going up but hasn't started going down yet. This means the vertical velocity is zero (\(v_y = 0\)).
2. Time of Flight (\(T\))
If a projectile is launched from the ground and lands on the ground, its total vertical displacement is zero (\(s_y = 0\)).
3. Horizontal Range (\(R\))
This is the total horizontal distance traveled. You find this by calculating the Time of Flight first, then multiplying it by the constant horizontal velocity:
\(R = u_x \times T\)
Encouragement: These formulas can look scary, but they just come from the basic SUVAT equations. You don't always need to memorize them if you know how to set up your SUVAT table!
5. The Cartesian Equation of the Trajectory
Sometimes we want to know the path of the projectile (\(y\) in terms of \(x\)) without caring about the time (\(t\)). By combining the horizontal and vertical equations, we get this (don't worry, you can derive this by substituting \(t = \frac{x}{u \cos \theta}\) into the vertical displacement equation):
\(y = x \tan \theta - \frac{gx^2}{2u^2 \cos^2 \theta}\)
This equation proves the path is a parabola. It is very useful if you are given a coordinate \((x, y)\) that the projectile must pass through, like a ball clearing a wall.
Summary and Quick Review
Before you move on to practice questions, check you've got these "Golden Rules" down:
- Gravity (\(g\)) is always \(9.8 \text{ ms}^{-2}\) and always points down.
- Weight is a force: \(W = mg\).
- Projectiles follow a parabolic path.
- Horizontal velocity stays constant throughout the whole flight.
- Vertical motion is just a standard SUVAT problem with \(a = -9.8\).
- At the highest point, the vertical velocity (\(v_y\)) is zero.
Don't forget: Mechanics is about practice. Start with simple vertical drops, move to horizontal launches, and soon you'll be handling complex angles with ease!