Projectile Motion and Air Resistance
Aim: To find the ideal angle with the lowest velocity to shoot a netball and explore the effects of air resistance on my result.
Introduction
As a self-identified nerd, there was only one part of the school day that I dreaded: the PE lessons. Now to every other athletically talented child out there it may have been fine to run for hours outside but to me there could not be anything worse than having to go swimming or play football. That is of course, before I experienced my first game of netball.
The ‘non-contact’ – read: violent – sport originates from England in the late 1800s and is highly competitive and popular in many of its colonies. Although slightly hindered by my height of a mere 161cm I am still passionate about the sport and especially about my position as a shooter. Having wormed my way into the thirds with a mere four months of shooting experience, although having played for several years prior, is nerve wracking as your team is depending on your amateur skills. Therefore, in an attempt to improve my shooting as well as simply out of curiosity I want to find the best angle for a player at a given height to shoot from which also requires the lowest amount of velocity.
In order to explore my problem, I have to use utilise projectile motion, “a projectile is an object in which the only force acting upon it is gravity”, and projectile motion is the movement of said object with no horizontal acceleration. Therefore, I will find and graph the trajectory of a netball throw from several different angles
The set up for this problem can be seen like this:
Post Height
305 cm
Ring Diameter
38 cm
Height of player
161 cm
Horizontal distance
141 cm
Vertical distance
≈144 cm (Calculate distance from ball to top of head)
Netball mass
425g
All calculations and solutions are completed under the assumption that a human can shoot at a precise angle with no error and without an opposition defending the shooter. They also do not take into account the time limit of 3 seconds in which a player can hold the ball for and will operate under the idea that air resistance is negligible (for now).
Modelling the trajectory of a netball – Excluding air resistance
In order to find angle with the lowest initial velocity I have decided to first map out the trajectory of the netball with different angles just to get an idea of the movement of the ball and have a clearer picture to work with. Therefore, I will choose 4 angles above 45° that I believe will work and find the gradient of the line from the angle along with the two coordinates that represent me (0,0 to begin with) and the post (1.14, 1.44 to begin with). From these I can construct a parabola that will outline the trajectory the netball without air resistance.
Finding initial velocity
We don’t have the time which the projectile has spent in the air therefore in order to model the trajectory and find the initial velocity we can use one of the SUVAT equations.
v2=u2+2as
Where
v = final velocity
u = initial velocity
a = acceleration
s = displacement
In these cases, v = 0 and a = 9.82 these two will always be constant.
Netball’s Trajectory
Gradient of the line:
m=tan(θ)
m = the gradient of the line
θ = the angle in degrees that a line makes with the x-axis in the positive direction
The quadratic to map the trajectory of the netball (sans air resistance):
y=ax2+bx+c
For now, c = 0 to simplify things although it will represent my height.
When launch angle θ is 60°
Gradient =tan (60) = √3
Initial launch (Linear) y= √3x
dydx= 3
dydxof y=ax2+bx+c is:
dydx=2ax+b
∴3=2ax+b
At (0,0) b= √3
y=ax2+3x+c
At (1.14, 1.44)
1.44=a(1.142)+3(1.14)
a= -0.41 (2 dp)
Overall quadratic
y= -0.41×2+3x+c
However we need to check the maximum point of this graph just to verify that the ball has passed through the top of the hoop.
dydx=2(-0.41)x+√3
In order to find the maximum point we must find the stationary point where
dydx = 0
0= 2(-0.41)x+√3
x=2.11
The maximum point is at x = 2.11, which is further than the position of the post (x = 1.14), therefore the netball would reach the posts coordinates before it reached its peak height. This would mean that the ball wouldn’t go into the goal and the initial shooting angle needs to be above 60°.
Using the method above I was able to test out 4 launch angles between 70 to 85 (in 5 increments) to reach the understanding that the launch angle with the lowest velocity must be below 70° as that had the shortest displacement and the maximum point was the closest to the post.
The trajectory of the projectile launched from different angles
I have grouped my findings together in the graphs below.
Graph with c = 1.61
In the graphs below I have shifted c to 1.61 as that represents my height in metres, this consequently also shifted up the coordinates of the post to (1.14, 3.05) which matches the real height of the netball posts. This only shifts the y intercepts to 1.61 and caused the graphs to move upwards
Calculating initial velocity
Using the SUVAT equation that was introduced earlier we can then calculate the initial velocity for all four angles
v2=u2+2as
The overall initial velocity of the ball can be calculated with either the horizontal or vertical initial velocity as well as the initial launch angle with the following equations:
v=vxcos(θ)
v=vysin(θ)
Where:
vx=initial horizontal velocity
For 70°
0=(uy)2+2(-9.82)(1.80)
uy= -2(-9.82)(1.80)
uy=5.95 m/s
v=5.95sin(70)
v=6.33 m/s
75°
v=6.21 m/s
80°
v=6.58 m/s
From the previous 3 angles we can already infer that the relationship between the angle of launch, initial vertical velocity and initial velocity is not linear but rather more like a quadratic. Therefore the angle with the minimum velocity is between 70 and 75
From the 3 points we can create a parabola to represent the relationship between the angles, initial vertical velocity and initial velocity which is:
y= 0.0098×2-1.45x+59.46
From this we can differentiate and find the minimum point that will give us the angle with the lowest velocity.
dydx= 2(0.0098) x – 1.45
0= 2(0.0098) x – 1.45
x=73.72
y= 0.0098(73.72)2-1.45(73.72)+59.46
y=5.83
Therefore, the ideal angle at which to shoot a netball with the smallest amount of velocity is 73.72° where the velocity needed is 5.83m/s
Exploring ideas of air resistance and drag
Although we have found the ideal angle at which to shoot a netball that will pass through the coordinates of the hoop with the lowest velocity, this is only a hypothetical situation. Realistically we need to take into account air resistance as well as drag. In the absence of air resistance, the horizontal velocity of a projectile will remain constant as it accelerates downwards due to gravity. Therefore, if air resistance is present we will need to take into account effects of changing horizontal velocity, which may alter the ideal angle of shooting. There are also other certain physical restraints that we have to take into account.
When adding air force and drag onto the motion of the netball we will restrict our explorations of it to air resistance being proportional to velocity or velocity squared. However, air resistance and drag can be altered by several factors; the uniform velocity of the object through different fluids (Galileo’s principle of relativity), the shape and size of the object, the density of the fluid as well as other factors like the object’s surface texture. All of these factors will be represented by one variable (b).
F=bv2
F=bv
F = air resistance force
b = (surface texture, shape, density etc.)
v = velocity
As the netball is a relatively small object and is quite slow the drag force would be proportional to only velocity instead of velocity squared.
We can use the equations of motion in order to explore the effect of air resistance on the ideal angle of 73.72°.
The equation of motion is ma=mg-bvor mdvdt=mg-bv
v = velocity
m = mass of the netball
g = acceleration due to gravity
b = positive constant
This is from the equation F=mafrom Newton's second law of motion to refer to forces due to any acceleration and F= gmwhich is used specifically to define an object’s weight and the effects of gravitational acceleration.
We can separate the equation into one for the x component and y component
dvxdt= -gvxvt
dvy dt =-g(1+vyvt)
Here the velocity of the netball (v) = (vx, vy)
At this point vt=mgb as it is at terminal velocity where the air resistance is equivalent to the gravitational force.
We can integrate this to get
v0xvxdvxvx= -gvtt
Where v0x= v0cosθ
Therefore
ln (vxv0x) =-gvtt
From this equation we can tell that air resistance causes the horizontal velocity to decrease exponentially, therefore the overall velocity will also decrease, as the angle of launch remains constant. Thus, the maximum point of the ball’s trajectory will fall at a greater rate than the increase of drag force upon the ball.
Evaluation and application
After finding the optimum angle I should shoot a netball while using the lowest velocity I realised I should put this into effect and test my solution. This obviously proved to be difficult to implement as I cannot judge the angle I shoot within 3 seconds before I lose the ball due to holding. Even without the time constraint my judgement of my angle of throw is inaccurate as it lacks precision, I cannot judge by sight the angle of my launch. Therefore, although I have found a mathematical solution it will be quite hard to apply it to a real-world situation. Furthermore, my exploration does not take into account the presence of an opposition, as depending on their height I may not be able to shoot at the ideal angle. My actual aim will also affect either the ball gets in the hoop or not and the mass of a netball can vary between 400g and 450g.
However the idea of
Conclusion
In the end I am very happy to have reached a final solution for my aim: I have found the angle at which a person of given height (1.61m) should shoot a netball into a post (3.05m) with the minimum amount of velocity.
However, although I have reached my aim there is still many limitations as it does not take many factors into account. Having played in several netball matches throughout the duration of my exploration I have also realised that during a game and under time pressure a player can only trust their instincts, there is no time to adjust the launch angle to a specific value, therefore although my research was successful it may not be effective in real life situations, even after evaluating the effect of air resistance on the motion of the ball.
Although I have managed to solve my original aim of finding an ideal angle with the lowest velocity it will be more difficult to implement it in real life. My evaluation of air resistance can also be more detailed and I could explore the effects of wind or rain on the motion of the ball if I were to extend my study. After my research I have gain a greater appreciation for the application of maths to real life problems although they do not always provide a realistic solution to the problem due to human complication. I have also learnt that the link between velocity and the angle of launch is a quadratic and the shortest displacement does not equal the lowest overall velocity.
In the end it doesn’t matter that the angle I have found may not be implemented into my every day play, just the exploration has caused me to practice my shooting more frequently. Overall, I have improved steadily and my percentage of success is rising.