Projectile Motion 2

Projectile Motion 2
In this post, I will expand on my previous one with things relating to the range, time and distance. Expanding on from before:

We will now derive an equation for the time of flight, from leaving the ground to hitting it. Consider the case below:

At the peak height (very top) the vertical velocity is 0 so:

From this we can see the time to the peak height is given by the formula above, now we can find the total time for the ball to reach the ground. If your intuition is right, you should have realised you need to multiply the formula above by 2. This is because the trajectory is a parabola (we will go more into that in part 3) and so symmetric about the peak; also because of the conservation of energy the velocity going up and the velocity going down will be equal and opposite for a certain height (as illustrated by the example at height H in the above diagram). As an extension see if you can show that fact from the equation at the bottom of the post. So:

(where T is the time to hit the ground, note this is where the ground is level)

From this, we can now calculate the maximum horizontal distance (the range) by using our formula for the constant horizontal velocity, multiplying the horizontal velocity by the total time of the flight you get the range as:

(Here I simplified the formula for range by using trigonometric identities, don't worry too much)

Now we can see to achieve the maximum horizontal range we need an angle of 45 degrees, as we multiply this by 2 to get 90 degrees which is where sine is at its peak value. This makes sense if you think about it.

To find the peak height we can use the conservation of energy:

(In this example capital V is the initial velocity and the lowercase v is the velocity at a certain height )

(I got the second equation by cancelling the mass and multiplying each term by 2)

The statement above comes from the fact that energy cannot be created nor destroyed and says that at any given point along the trajectory the sum of the objects kinetic and gravitational energy stores will equal the total kinetic energy store of the ball to begin with (the kinetic energy it had at the very beginning). Now, at the peak height, all the velocity is in the horizontal direction which we know from before is VcosO, and so rearranging and substituting gives:

(here I used the Pythagorean identity that cosx squared plus sinx squared equals 1, this is trivial to prove)

Now we have the formula for the peak height, as you can see to achieve the maximum height possible we need an angle of 90 degrees with the ground which should also be obvious when you think about it.

Thank you for reading part 2, especially if you struggled with part 1, and I hope you understood some if not all parts of this, and look out for part 3-Joe

Projectile Motion 1

I understand some of you dislike maths, so well done if you are one of those who are reading this. Most people were reluctant to study science and maths even when they had to, therefore I don't expect many to be interested in this. This is more of an experiment for me to see how easy my content is to understand, so if it all goes over your head don't be discouraged and look out for other posts.
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).