Projectile Motion: Your Guide to Flying Objects!
Hey everyone! Ever wondered about the beautiful arc of a basketball heading for the hoop, a football soaring through the air, or even water from a fountain? That's projectile motion in action! In this chapter, we're going to learn the physics behind any object that is thrown or launched into the air.
The great news is that it's not as complicated as it looks. We'll break it down into simple, manageable pieces. By the end, you'll be able to predict the path, flight time, and landing spot of a projectile. Let's get started!
The Most Important Idea: Independence of Motion
This is the golden rule of projectile motion. If you understand this one concept, everything else will fall into place. Here it is:
The horizontal motion of a projectile and its vertical motion are completely independent of each other.
What does this mean? It means we can pretend we're looking at two separate objects: one moving only horizontally, and one moving only vertically. The only thing they share is time.
Analogy: The Dropped vs. Fired Bullet
Imagine a bullet is dropped from a certain height. At the exact same time, another bullet is fired horizontally from the exact same height. Which one hits the ground first? (Assuming a flat ground and no air resistance)
They hit the ground at the same time!
This sounds crazy, but it's true! The force of gravity pulls them both downwards at the same rate. The fired bullet's horizontal speed doesn't change its vertical journey at all. This perfectly shows the independence of horizontal and vertical motion.
Breaking Down the Motion
Let's look at the two "separate" motions in detail. For all our calculations, we will assume air resistance is negligible.
1. Horizontal Motion (x-direction)
- Force: There are NO horizontal forces acting on the projectile (we're ignoring air resistance).
- Acceleration: Since there's no net horizontal force (F=ma), the horizontal acceleration is zero ($$a_x = 0$$).
- Velocity: With zero acceleration, the horizontal velocity is constant throughout the entire flight. It never changes!
- Equation: The only equation you need is the simple one for constant velocity:
Distance = Speed × Time
$$s_x = u_x t$$
2. Vertical Motion (y-direction)
- Force: The only force acting is gravity, which pulls the object downwards.
- Acceleration: The vertical acceleration is constant and directed downwards. It's the acceleration due to gravity, g. We usually take $$g \approx 9.81 \, \text{m s}^{-2}$$.
- Velocity: The vertical velocity is constantly changing. It decreases on the way up, becomes zero at the very top, and increases on the way down.
- Equations: Since this is uniformly accelerated motion, we use our trusty equations of motion: $$v_y = u_y + a_y t$$ $$s_y = u_y t + \frac{1}{2} a_y t^2$$ $$v_y^2 = u_y^2 + 2 a_y s_y$$ (Remember to use vertical components only!)
Quick Review Box
Horizontal (x-motion)
- Acceleration $$a_x = 0$$
- Velocity is CONSTANT
- Use $$s_x = u_x t$$
Vertical (y-motion)
- Acceleration $$a_y = -g$$ (if 'up' is positive)
- Velocity CHANGES
- Use equations of motion
The bridge connecting them is TIME (t). It's the same for both.
Key Takeaway
To analyse projectile motion, we split the problem in two. We look at the horizontal part and the vertical part separately. Don't mix them up!
The Shape of the Path: A Parabola
So, what do you get when you combine a constant horizontal velocity with a constant vertical acceleration? You get a beautiful, symmetric curve called a parabola.
Think about the path of a basketball shot - it goes up and then comes down in a perfect arc. This curved trajectory is always a parabola, as long as we ignore air resistance.
Did you know?
Before the 17th century, people had many incorrect ideas about the paths of projectiles. It was the famous scientist Galileo Galilei who first correctly described the path as a parabola by using the idea of splitting motion into horizontal and vertical components.
Key Takeaway
The path a projectile takes through the air is called its trajectory, and it has the shape of a parabola.
How to Solve Projectile Motion Problems: A Step-by-Step Guide
Don't worry if the maths seems tricky at first. If you follow these steps every single time, you'll master these problems.
Step 1: Set up and Resolve
First, draw a quick diagram. Define your positive directions (e.g., right is +x, up is +y). If the object is launched with an initial speed u at an angle θ to the horizontal, you MUST resolve this velocity into its horizontal (x) and vertical (y) components.
Horizontal initial velocity: $$u_x = u \cos\theta$$
Vertical initial velocity: $$u_y = u \sin\theta$$
Memory Aid:
The horizontal component is 'close' to the angle θ, so it gets 'cos'.
Step 2: Create Two Lists (The T-Chart Method)
This is the most important step! Create a chart to separate your horizontal and vertical information. Let's use 'up' as the positive direction.
Horizontal (x)
$$s_x = ?$$
$$u_x = u \cos\theta$$
$$v_x = u_x$$
$$a_x = 0$$
$$t = ?$$
Vertical (y)
$$s_y = ?$$
$$u_y = u \sin\theta$$
$$v_y = ?$$
$$a_y = -g = -9.81 \, \text{m s}^{-2}$$
$$t = ?$$
Notice that time (t) is in both lists because it's the link between them!
Step 3: Find the Time of Flight
Most problems require you to find the total time the projectile is in the air. You will almost always use the vertical motion information to find this.
Example: If an object lands at the same height it started, its final vertical displacement $$s_y = 0$$. You can use $$s_y = u_y t + \frac{1}{2} a_y t^2$$ to solve for t.
Step 4: Solve for What's Asked
Once you have the time of flight (t), you can use it in either list to find other quantities.
- To find the horizontal range (how far it travelled horizontally), use the horizontal list: $$s_x = u_x t$$.
- To find the maximum height, use the vertical list. A key trick here is to remember that at the very peak of the trajectory, the vertical velocity is momentarily zero ($$v_y = 0$$).
Common Mistakes to Avoid!
- Forgetting to resolve the initial velocity: Never use the total speed `u` in your horizontal or vertical equations. Always use `u_x` and `u_y`.
- Mixing variables: Don't put `u_x` into a vertical equation, or `a_y` into a horizontal one! Keep them separate.
- Sign errors: Be consistent with your positive direction. If up is positive, then `g` must be negative (`a_y = -g`), and any displacement downwards is also negative.
- Thinking acceleration at the top is zero: The vertical velocity `v_y` is zero at the peak, but the acceleration is still `g` downwards! Gravity doesn't switch off.
Worked Example
A golf ball is hit from the ground with an initial velocity of 40 m s⁻¹ at an angle of 30° to the horizontal. Find:
(a) The time it takes to reach its maximum height.
(b) The maximum height reached.
(c) The total time of flight.
(d) The horizontal range of the ball.
(Take g = 9.81 m s⁻²)
Solution:
Step 1: Resolve the initial velocity.
$$u_x = u \cos\theta = 40 \cos(30^\circ) = 34.64 \, \text{m s}^{-1}$$ $$u_y = u \sin\theta = 40 \sin(30^\circ) = 20 \, \text{m s}^{-1}$$Step 2: Set up the lists (Let's choose 'up' as positive).
Horizontal (x)
$$s_x = \text{Range} = ?$$
$$u_x = 34.64$$
$$a_x = 0$$
Vertical (y)
$$s_y = \text{Max Height} = ?$$
$$u_y = 20$$
$$a_y = -9.81$$
(a) Time to reach maximum height
At the maximum height, $$v_y = 0$$. Let's use the vertical list.
$$v_y = u_y + a_y t$$ $$0 = 20 + (-9.81) t$$ $$9.81 t = 20$$ $$t = \frac{20}{9.81} = 2.04 \, \text{s}$$(b) Maximum height
We can use the time we just found (t = 2.04 s) or another vertical equation. Let's use:
$$v_y^2 = u_y^2 + 2 a_y s_y$$ $$0^2 = 20^2 + 2(-9.81)s_y$$ $$0 = 400 - 19.62 s_y$$ $$19.62 s_y = 400$$ $$s_y = \frac{400}{19.62} = 20.4 \, \text{m}$$(c) Total time of flight
The path is symmetrical. The time to go up is the same as the time to come down. So, the total time is twice the time to reach the maximum height.
$$\text{Total time} = 2 \times 2.04 = 4.08 \, \text{s}$$(d) The horizontal range
Now we use the horizontal list with the total time of flight (t = 4.08 s).
$$s_x = u_x t$$ $$s_x = 34.64 \times 4.08 = 141.3 \, \text{m}$$Key Takeaway
The problem-solving method is always the same: Resolve the initial velocity, Separate the information into horizontal and vertical lists, and then Solve using the correct equations for each part, with time as the link.
Final Summary: You've Got This!
Projectile motion might seem intimidating, but it all boils down to three simple ideas:
- Independence of Motion: Horizontal motion is constant velocity ($$a_x=0$$). Vertical motion is constant acceleration ($$a_y=-g$$). Treat them separately!
- Parabolic Path: The trajectory is always a parabola (if we ignore air resistance).
- Time is the Key: Time (t) is the one variable that is the same for both horizontal and vertical motions, linking them together.
Keep practising by following the step-by-step method, and you'll become confident in solving any projectile motion problem. Good luck!