Mechanics Coursework

In order to model the basketball as a particle I have made the following assumptions:

• That there is no air resistance.

I have made this assumption because the effect of air resistance on the speed of the basketball would make it extremely difficult to model.

• That gravity is 9.8ms-2.

I have made this assumption because the strength of gravity on earth is <blah.blah> and I have taken this as being 9.8 as 2 significant figures is sufficient for accuracy. Obviously it is sensible to use Earth as the planet o choice. It would be interesting to investigate the difficulties of futuristic ‘moon basketball’ to take the investigation further.

• I am also assuming that the basketball does not ‘spin’.

I am making this assumption because the additionally force of ‘spin’ would be an extra complication to the model which would increase the difficulty in trying to model the motion of the basketball.

• I am also making the assumption that the ball goes through the centre of the hoop, without hitting the backboard.

This is because the additional forces involved if the ball hit the rim of the hoop or the backboard would complicate matters as the path the basketball takes would no longer be a parabola.

• I am assuming that the ball travels at a constant speed

To simplify the model

• I am ignoring the effects of wind

To keep the model simple

• That the basketball hoop 3.05m high, the player 1.98m high, and the distance of the centre of the hoop from the free-throw line is 4.22m.

I am using these values because they were the ones I found through research and are consistent with my own knowledge. Also the height of the player shall stay constant as I cannot have more than one variable.

• I am taking the speed a person can throw a ball to be 10ms-1

This is based on an experiment I did myself where I threw a ball over a distance of 10m and had the time I took recorded by another person

The results were as follows:

By the equation speed = distance/time an estimate of the speed a person can throw a ball can be found:

Speed = 10/1.10

Speed = 9.10ms-1

An estimate of the speed will need to be constantly used in the models as we cannot vary this as I have chosen to vary the angle.

I am now going to find mathematical solutions for the optimum angle needed to score a basket in basketball using constant acceleration equations. I am assuming ax = 0ms-1 and ay = -9.8ms-1. sx = 4.22m. sy = height of basketball hoop – height of player.

sy = 3.05 – 1.98

sy = 1.07m

The five constant acceleration equations are:

v = u + at

s = ½ (u + v)t

s = ut + ½ at2

s = vt – ½ at2

v2 = u2 + 2as

Variables are:

u, initial velocity, ms-1

v, final velocity, ms-1

t, time, s

All my measurements are accurate to 3 significant figures

Analysing my Model

The initial angles I shall use in my equations will be: 20°, 30°, 40°, 50°, 60°, 70°, and 80° degrees.

Once I have used these angles to discover whether the basketball falls through the hoop or not I shall narrow the range of angles e.g. 50°, 55°, 60°.

Model

When a particle moves in projectile motion its displacement and velocity are written as components.

The displacement of the particle in the x and y direction can be found from the constant acceleration equation

s = ut + ½ at2

When we use this equation solely to find the displacement in the y direction it is written

y = utsinθ – ½ gt2 

This is because initial velocity (u) in the y direction is usinθ, and acceleration (a) in the y direction is – gravity (g).

We can see why u in the y direction is usinθ if a velocity triangle is drawn:

When we use the equation s = ut + ½ at2 solely to find the displacement in the x direction it is written

x = utcosθ

This is because initial velocity (u) in the x direction is ucosθ (based on the above velocity triangle), and acceleration (a) in the x direction is 0 as we are ignoring any effects such as wind which could cause horizontal acceleration.

The velocity components of the particle are found using the constant acceleration equation

v = u + at

When we use this equation to find the velocity in the y direction it is written

vy = usinθ – gt

When we use the equation v = u + at to find the velocity in the x direction it is written

vx = ucosθ

To summarise:

The next step in my investigation is to isolate the variables u, v and t (initial velocity, final velocity and time) onto one side of an equation so their values can be found.

This can be done through substitution: where one equation is substituted into another.

Firstly, I shall form an equation that isolates the initial velocity (u).

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