Projectile Motion 1
I hope for this to be a simple post, however, I am unsure as to how some people who struggle with maths may find it, so I intend to split it over 2 or 3 posts with varying difficulty.
This topic appears in A-Level physics and A-Level maths
Projectile motion describes how an object launched behaves in flight and describes its corresponding trajectory: for example throwing a stone or kicking a ball. In its simplest case, we can look at how a ball that is launched with some angle and speed moves over the ground.
To begin with:
The Black object is our ball and the blue vector is the velocity of the ball at the very start. As it is a vector quantity it will have a magnitude (the speed of the ball) and a direction (the angle with the surface). In this example, we will be ignoring drag, but before it is important to go over vectors:
Our blue velocity vector can be split apart (resolved) into an upward speed and a side speed, and adding these two together will result in the blue velocity vector, which we gave the ball. The upward speed is our vertical velocity at the very start and the side speed is our horizontal velocity at the very start. If we know the angle separating v from the ground, we can then calculate both of these velocities using trigonometry:
Now we need to take into account gravity: since gravity is only acting in the downwards vertical direction it will have no effect on the horizontal velocity, but over time it will reduce the vertical velocity. We know the acceleration due to gravity is:
(it's negative because it is downwards and the triangle means change in)
From this:
I understand if you may be a little confused but all I have done here is rearranged the formula; remember that the change in time is just the time since you kicked the ball (t) with time starting at 0, and the change in vertical velocity being the vertical velocity after a certain time minus the starting vertical velocity, which is VsinO. So now:
(where the vertical velocity is given as a function of time, the starting angle and the starting speed)
We can do the same for the horizontal velocity, however since gravity only acts vertically the horizontal velocity will be a constant and it won't change, so it will just be what I calculated to begin with:
Now we have these two equations that describe the horizontal and vertical velocity of our ball after a certain time t:
I used the Pythagorean Theorem to calculate the magnitude of the velocity tangent to the trajectory as the vertical and horizontal components form a right-angled triangle. Don't confuse this with the vector V.
As you can see from the two equations above describing the components of the velocity, the horizontal velocity is constant throughout, and the vertical velocity is reduced by 9.81m/s every second. These two equations can be used to find more information about the trajectory, such as maximum range, speed, peak height, and duration.
If you understood this well and enjoyed it please read the next posts, else don't worry and try to re-read it in case you missed something. Often not understanding something in maths comes from having shaky foundations in the things before. I hope you enjoyed reading and 'thank you for wanting to know more today than you did yesterday'(Neil deGrasse Tyson).
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