1. MMPS Project 1: The Throwing Problem
Group 6: Cameron M, Curtis C, Henry S, Melissa P, Sabina A, Axel H
October 2015
1 The Problem
During Fresher’s week, and to coincide with the 2015 Rugby World Cup, one
student makes a bet with their housemates that they are able to throw a rugby
ball over the roof of their house, run through the house, and then catch the ball
as it is on its way down in the garden. Before accepting the bet, the housemates
ask your advice as to whether this is really feasible. Is it?
To answer this problem we need to first show if it is possible using an optimal
scenario taking various assumptions. Then once we’ve established whether it is
possible in an optimal scenario we can use more appropriate values taking less
assumptions and see if it is feasible in a realistic scenario.
2 The Optimal Scenario
1. The Assumptions:
Trajectory Of Throw:
For our model to be correct we must make the assumption that the flat-
mate throws in a 2D plane and there is no wind pushing the ball off its
path, this will allow us to model the problem in 2D and make it much
simpler.
Gravity:
We will assume the acceleration due to gravity on the ball to be −9.8ms−1
and denote it as g. This is a standard value and will be used in all
calculations.
Path Through House:
We will assume that the flatmate will have a direct, straight path through
the house from the front door to the back door and also that these doors
will be opened prior to the throw.
Shape Of House:
We have assumed the house to be a symmetrical 2 floor house with walls of
height wm, and additionally a sloped roof adding a maximum additional
height of ym in the center of the house, giving a maximum height of the
house, denoted as zm. Another assumption we have made is the roof and
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2. house is clear of any obstructions such as a chimney, guttering or satellite
dish.
Ability Of the Flatmate:
We have assumed our flatmate to be 18 years old and therefore will use
average values of an 18 year old for height, speed they can throw the ball
and run. We’ve assumed they will throw the ball at 2/3 the speed of a
professionals value of roughly 60mph. Then again at 2/3 the speed for
running, which for a professional is roughly 10.6ms−1
.
Air Resistance: A rugby ball is quite wide which means that it cannot be
thrown very far, but can be thrown quite accurately. The person throwing
the ball must decide on the angle and the distance they throw it depending
on the wind present. If the wind is coming strongly from behind, the ball
should be thrown at a steeper angle and more closer to the house since
as the ball goes higher up, the wind will propel the ball further. This is
because the wind does not have any obstructions at higher altitudes and
therefore it exerts more effect on the ball. However if the wind is coming
towards the person, the ball should be thrown at a smaller angle and fur-
ther back from the house. Due to the fact of there being stronger winds
towards them, the ball will go further if it is kept low and this will only
work if the person is standing further away in order to still get it over the
roof.
Furthermore, the velocity of the ball when it is thrown must be at a high
speed for it to actually travel the length of the house. When the ball is
thrown upwards the vertical velocity must be high and for it to go across
the house it must have a high horizontal velocity also. The deceleration
due to gravity will bring the vertical velocity to zero and this will bring
the ball down to the ground in the shape of a parabolic curve. This is
why it is important to have a very high initial velocity so that that the
ball doesn’t descend too quickly onto the house rather than clearing it.
The person can also consider throwing the ball with a spiral motion in
order to increase the distance it travels. Spiralling is when the ball rotates
at a high speed along it’s horizontal axis due to the manner in which it’s
thrown. This method will work better because as the ball spirals, it will
reduce the air resistance it experiences which will increase the horizon-
tal velocity as well as the distance covered. The rugby ball will become
more stable which will allow for a smooth throw and make it easier for
the flatmate to catch it at the end. In our scenario we have decided not
to include a variable of air resistance as we feel there is no specific mathe-
matical function we can integrate into to allow us to accurately gauge the
effect of air resistance on our ball and throw. We feel this is reasonable as
to tackle this problem we are considering an ideal scenario.
Width of the Ball: The width of the rugby ball is 9.55cm which when added
into the equations would give a near negligible difference of final value. To
be exact, the difference in time in the air would be 3.57 seconds compared
to 3.61 seconds which we considered to be almost incomprehensible to the
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3. action taking place.
2. The Values:
Height of House (w): 6.1m
Height of Roof (y): 1.9m
Total height of house + roof (z): 8.0m
Height off the ground from which the ball is thrown (a): 2.5m
Distance to run within house (L): 8.7m
Speed at which flatmate can run: 7.1ms−1
Speed at which flatmate can throw the ball (V): 17.9ms−1
3. The Model
4. The Maths:
Begin by taking Distance
T ime = V elocity and by putting in our notation we
get the equation:
L + 2x
T
= V cosθ
Rearranging this formula can give us an expression for T:
T =
L + 2x
V cosθ
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4. Before we can go any further we need to find some equations that will give
us expressions for T. We can do this by deriving some kinematic equations
which are referred to in shorthand as S.U.V.A.T. Firstly:
v = u + at
Acceleration is defined as change of velocity over time. a = v−u
t which we
can rearrange easily to give us v = u + at.
s = ut +
at2
2
First we must derive s = u+v
2 t. This rearranges to give v = s
t from which
we get the expression s = vt. The average velocity is v+u
2 . Substituting
this in for v gives s = v+u
2 t.
Using the two previously derived equations will give us another expression:
s = 2u+at
2 t which we then simplify to give s = ut + at2
2 . Now these
expressions are ready to be used in our working.
We can solve the earlier expression using the S.U.V.A.T equation v = u+at
to get another expression for T. The lowercase t in this case denoting the
time taken for the ball to make the horizontal journey.
−V sinθ = V sinθ − 9.8T
Now this equation can be easily rearranged to find another expression for
T, as shown here:
T =
2V sinθ
9.8
As these are both expressions for T we can equate them to give the ex-
pression:
L + 2x
V cosθ
=
2V sinθ
9.8
This can then be rearranged to get an expression for x by multiplying
through by V cosθ and doing simple rearrangements, to finally give us an
expression we can name
(∗) : x =
5
49
V 2
sinθcosθ −
L
2
Since we know that Distance
T ime = V elocity gives us another equation when
we substitute in the algebraic equivalents, such that:
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5. x
t
= V cosθ ⇒ t =
x
V cosθ
Now we’ll use another kinematic formula to get a quadratic equation which
we can then use to find a suitable value for the optimal angle to throw,
by simply inputting values for a distance to run and speed at which the
ball is thrown. The kinematic equation we will use is S = ut + at2
2 where
S denotes the height the bal has to travel before clearing the house’s
guttering, being measured as S = w − a. Then first putting the initial
values in we come to get the equation:
S = (V sinθ)(
x
V cosθ
) −
49
10
(
x
V cosθ
)2
Now we can substitute the expression for x we have from (∗) to get the
equation:
S =
5
49 V 3
sin2
θcosθ − LV sinθ
2
V cosθ
−
49
10
(
25
2401 V 4
sin2
θcos2
θ − 5
49 V 2
Lsinθcosθ + L2
4
V 2cos2θ
)
As you can see that equation is very lengthy. Hence, with quite a sub-
stantial amount of expanding out and then simplifying, we eventually get
the equation:
(
5V 2
98
)n2
+ (−S −
5V 2
98
)n + (S −
49L2
40V 2
) = 0
Now we can input our values for L and V we get a value that we have
denoted as n where n = sin2
θ, from which we can obtain the optimal
angle for which the ball should be thrown in order to both maximise the
length of time the ball is in the air and the best arch where it will fall
closest to the house thus minimising the distance the flatmate has to run.
We will also use this angle to work out the time the ball will be in the
air and the distance the flatmate needs to stand back from the house to
achieve this perfect arch.
So when we put in our values we get:
1602.05
98
n2
+ (−3.6 −
1602.05
98
)n + (3.6 −
3708.81
12816.4
) = 0
We can then take this quadratic and use the quadratic formula to solve
for a value of n
−b ±
√
b2 − 4ac
2a
⇒
12.747 ± 162.497 − 4(16.347)(3.311)
32.695
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6. The quadratic equation above gives solution of n = −0.447. This then
works out to give the value for θ = 81.2°which we can then put back into
our earlier equations to find T and x.
x = 32.695(sin(81.2))(cos(81.2)) − 4.35 ⇒ x = 0.593m
T = 3.653sin(81.2) ⇒ T = 3.61seconds
Now if we compare this time against the time it would take the flatmate
to run the full distance L + 2x ⇔ 8.7 + 2(0.593) ⇔ 9.886m we can see if
it is feasible. The speed at which our flatmate runs is 7.1ms−1
and the
distance to cover is 9.886m, so they will cover that distance in 9.886
7.1 = 1.392
seconds, thus having plenty of time to catch the ball.
3 Conclusion
The problem poses the question of is the act physically feasible?
In the case of our optimal scenario, the answer to this is clearly Yes since the
flatmate has almost 2 full seconds after they have covered the required distance
on the ground before the ball reaches the height of their outstretched arm.
Furthermore, of course the flatmate can still catch the ball at any lower height
before it reaches the ground.
While our model may seem simplistic, and it may have been ideal to construct a
further scenario with increased variables and a more accurate model, ours fully
answers the question posed with enough mathematical evidence to state it as
correct and justified.
4 Bibliography
Effect of air resistance on rugby ball: http://envisionrugby.weebly.com/
Speed of ball throw: http://ftw.usatoday.com/2014/03/how-fast-football-
throw-nfl-combine-logan-Thomas
http://www.cbssports.com/nfl/eye-on-football/24466665/colin-kaepernick-no-longer-
has-nfl-combine-record-for-fastest-pass
Average Height of 18 year old: http://www.bbc.co.uk/news/uk-11534042
Speed of running: http://voices.nationalgeographic.com/2013/08/09/how-
fast-can-a-human-run/
Values for House Averages: http://reneweconomy.com.au/2013/how-big-is-
a-house-average-house-size-by-country-78685
We took the value for the UK on that site and square rooted them in order to
get a width of the house.
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