Question 2
Topic – AHL 3.12
A sports stadium has a T-shirt cannon which is used to launch T-shirts into the crowd. The purpose of this question is to determine whether a person sitting in a particular seat will ever receive a T-shirt.
A T-shirt cannon is placed on the horizontal ground of a stadium playing area. A coordinate system is created such that the origin, O, is the point on the ground from where the T-shirts are launched. In this coordinate system, $x$ and $y$ represent the horizontal and vertical displacement from O, and are measured in metres.
Seat $A_1$ is the nearest seat to the T-shirt cannon. The coordinates of the front of the foot space for seat $A_1$ are $(30, 2.1)$.
Each seat behind seat A 1 is 1.0 m further from O horizontally and 0.5 m higher than the seat in the row below it, as shown on the diagram.
Seat $A_1$ is in row 1. Let seat $A_n$ be the seat directly behind $A_1$ in row $n$.
(a) (i) Write down the coordinates of the front of the foot space of seat $A_5$.
(ii) Find, in terms of $n$, the coordinates for the front of the foot space of seat $A_n$.
While in motion, the T-shirt can be treated as a projectile.
Let $t$ be the time, in seconds, after a T-shirt is launched.
At any time $t > 0$, the acceleration of the T-shirt, in $ms^{-2}$, is given by the vector
\[\binom{\ddot{x}}{\ddot{y}} = \binom{0}{-9.8}.\]
The initial velocity, in $ms^{-1}$, of the T-shirt is given as
\[\binom{29.4\cos\theta}{29.4\sin\theta},\]
where $\theta$ is the angle to the ground at which the T-shirt is launched and $0^\circ < \theta \leq 90^\circ$.
(b) (i) Find an expression for the velocity, $\binom{\dot{x}}{\dot{y}}$, at time $t$.
(ii) Hence show that when the T-shirt is launched vertically, the time for it to reach its maximum height is 3 seconds.
The displacement of the T-shirt, $t$ seconds after it is launched, is given by the vector equation
\[\binom{x}{y} = \binom{29.4(\cos\theta)t}{29.4(\sin\theta)t – 4.9t^2}.\]
(c) Using the given answer to part (b)(ii) or otherwise, find the maximum height reached by a T-shirt when it is launched vertically.
(d) (i) If there was no seating, and the T-shirt was launched at an angle $\theta$, show that the value of $x$ when it would hit the ground is given by the expression
\[x = 176.4\sin\theta\cos\theta.\]
(ii) Hence find the maximum possible value for $x$ if there was no seating to block the path of the T-shirt.
In order to calculate the seats in the stadium which can be reached by a T-shirt it is required to find the equation of the curve that forms the boundary of all the points that can be reached. This boundary is represented by the dashed curve in the following diagram, while the solid curves represent some of the possible trajectories for the T-shirts.
It is given that the boundary curve is the parabola $y = ax^2 + bx + c$, with its vertex $V$ on the $y$-axis.
(e) Using your answers to parts (c) and (d)(ii), or otherwise, find
(i) the value of $c$.
(ii) the value of $b$.
(iii) the value of $a$.
A spectator is sitting in seat $A_{40}$.
(f) Show that it is not possible for the spectator to ever get a T-shirt.
▶️Answer/Explanation
Solution: –
(a) (i) (34, 4.1)
(ii) recognizing the sequence is arithmetic, with a
common difference of 0.5 and 1.0
\[(30+(n-1),2.1+(n-1)0.5)\]
\[(=(29+n,1.6+0.5n))\]
(b) (i) evidence of integration of the acceleration vector OR use of v = u + at
\[\binom{\dot{x}}{\dot{y}} = \binom{c_{1}}{-9.8t+c_{2}}\]
\[\binom{x}{y} = \binom{29.4\cos\theta}{29.4\sin\theta-9.8t}\]
(ii) $\theta=90^{\circ}$
$29.4-9.8t=0$
maximum point when $t=\frac{29.4}{9.8}$
= 3 (seconds)
(c) correct substitution OR use of correct graph
maximum height is $29.4 \times 3 – 4.9 \times 3^{2}$
= 44.1 (m)
(d) (i)
\[29.4~sin~\theta t-4.9t^{2}=0\]
t=6sine (or \(t=0)\)
\[x=29.4~cos~\theta\times6~sin~\theta\]
\[=176.4~cos~\theta~sin~\theta\]
(ii) valid method to find maximum (e.g. sketch graph, find derivative)
maximum value of x is 88.2 (m)
(e) (i) \((c=)\) 44.1
(ii)
EITHER
\[\frac{dy}{dx}=0\Rightarrow2a\times0+b=0\]
OR
vertex is at \(x=-\frac{b}{2a}=0\)
THEN
\[\Rightarrow b=0\]
(iii) point (88.2.0) used
\[0=a\times88.2^{2}+44.1\]
\[\Rightarrow a=-\frac{44.1}{88.2^{2}}\]
\[=-\frac{5}{882}=-0.0056689.\]
(f) \$use of their arithmetic sequence from (a)(ii) with n=40\$
\$coordinates of seat \(A_{40}=(69,21.6)\)\$
$EITHER$
\$substitution of their 69 into their \(y=-\frac{5}{882}x^{2}+44.1\) from part (e)\$
\[y=-\frac{5}{882}\times69^{2}+44.1\]
=17.1 (17.1102…)
\[21.6>17.1\]
$OR$
\$substitution of their \(y=21.6\) into their \(y=-\frac{5}{882}x^{2}+44.1\) from part (e)\$
\[21.6=-\frac{5}{882}x^{2}+44.1\]
\[x=63\]
\[y=-\frac{5}{882}x^{2}+44.1\]
from part (e)
\[69>63\]
So the T-shirt cannot reach seat \(A_{40}\)