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Mechanics Coursework

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Mechanics Coursework

Scoring a Basket in Basketball

Formulating my Model

My task is to produce a strategy necessary for scoring a basket in basketball. I shall investigate the effects of throwing the ball at different angles and ascertain the ideal angle for scoring a basket. In addition, I shall investigate what would be the best angle for me to throw the basketball rather than just the basketball player as the angle will be different as I am much shorter than a basketball player. I will model the motion of the ball as it leaves the hands of the basketball player and falls through the hoop. I will model the basketball as a particle.

Basketball is a ball game where if a free throw is taken a player will try and shoot a hoop from the free-throw line which is 4.6[1]m from the backboard of the hoop. The centre of the hoop is then 38[2]cm from the backboard. Therefore, the centre of the hoop is 4.22m from the free-throw line.

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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°.




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

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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:

Vertical Displacement

Horizontal Displacement

s = ut + ½ at2

s = ut + ½ at2

y = utsinθ  ½ gt2

x = utcosθ

Vertical Velocity

Horizontal Velocity

v = u + at

v = u + at

vy = usinθ gt

vx = ucosθ

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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[4] This was an average based on the heights of one basketball team found on the website http://www.tigers.com.au/articles/nbl/018.php

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