IB Mathematics AHL 3.12 Vector applications to kinematics-AI HL Paper 2- Exam Style Questions- New Syllabus

Question

At an archery tournament, a ball is launched into the air, and an archer attempts to hit it with an arrow. The ball’s path is modeled parametrically, incorporating initial velocity components and gravitational effects, while the arrow follows a linear trajectory.

(a) (i) The path of a ball is modeled by \(\binom{x}{y} = \binom{5}{0} + t\binom{u_x}{u_y – 5t}\), where \(x\) and \(y\) are horizontal and vertical displacements (in meters), \(t\) is time (in seconds), and \(u_x = 8\), \(u_y = 10\) are the initial velocity components. Find the initial speed of the ball:
(ii) Find the angle of elevation of the ball as it is launched:

(b) Find the maximum height reached by the ball:

(c) Assuming the ground is horizontal and the ball is not hit, find the \(x\)-coordinate where the ball lands:

(d) For the path of the ball, find an expression for \(y\) in terms of \(x\):

(e) An arrow is released from (0, 2) with speed 60 ms\(^{-1}\) at 10° elevation. Determine the two positions where the arrow intersects the ball’s path:

(f) Determine the time when the arrow should be released to hit the ball before its maximum height:

▶️ Answer/Explanation

Markscheme

(a)(i)
    Velocity vector: \(\binom{u_x}{u_y} = \binom{8}{10}\)
    Initial speed = \(\sqrt{u_x^2 + u_y^2}\)
    Calculation: \(\sqrt{8^2 + 10^2} = \sqrt{64 + 100} = \sqrt{164} \approx 12.8062\)
Result:
12.8 ms\(^{-1}\)

(a)(ii)
    Angle: \(\tan \theta = \frac{u_y}{u_x} = \frac{10}{8} = 1.25\)
    Compute: \(\theta = \tan^{-1}(1.25) \approx 0.896\) radians
    Degrees: \(\theta = \frac{0.896 \cdot 180}{\pi} \approx 51.34^\circ\)
Result:
0.896 rad or 51.3°

(b)
    Position: \(\binom{x}{y} = \binom{5}{0} + t\binom{8}{10 – 5t}\), so \(y = 10t – 5t^2\)
    Derivative: \(\frac{dy}{dt} = 10 – 10t\)
    Maximize: Set \(\frac{dy}{dt} = 0\): \(10 – 10t = 0 \Rightarrow t = 1\)
    Height: \(y = 1 \times (10 – 5 \times 1) = 5\)
Result:
5 m

(c)
    Landing condition: \(y = 0\)
    Equation: \(y = t \times (10 – 5t) = 0\)
    Solve: \(t = 0\) (start) or \(10 – 5t = 0 \Rightarrow t = 2\)
    X-coordinate: \(x = 5 + 8 \times t = 5 + 8 \times 2 = 5 + 16 = 21\)
Result:
21 m

(d)
    X-equation: \(x = 5 + 8t\), so \(t = \frac{x – 5}{8}\)
    Y-equation: \(y = 10t – 5t^2\)
    Substitute: \(y = 10 \left(\frac{x – 5}{8}\right) – 5 \left(\frac{x – 5}{8}\right)^2\)
    Simplify: \(y = \frac{10(x – 5)}{8} – \frac{5(x – 5)^2}{64} = \frac{5(x – 5)}{4} – \frac{5(x^2 – 10x + 25)}{64}\)
    Further: \(y = -\frac{5}{64}x^2 + \frac{65}{32}x – \frac{525}{64}\)
    Factored form: \(y = -\frac{5}{64}(x – 5)(x – 21)\)
Result:
\(y = -\frac{5}{64}(x – 5)(x – 21)\)

(e)
    Arrow equation: Slope = \(\tan 10^\circ \approx 0.176\), \(y = 0.176x + 2\)
    Ball equation: \(y = -\frac{5}{64}(x – 5)(x – 21)\)
    Set equal: \(0.176x + 2 = -\frac{5}{64}(x^2 – 26x + 105)\)
    Solve: \(64(0.176x + 2) = -5(x^2 – 26x + 105)\), \(11.264x + 128 = -5x^2 + 130x – 525\), \(5x^2 – 118.736x + 653 = 0\)
    Quadratic roots: Discriminant = \(118.736^2 – 4 \times 5 \times 653 \approx 1038.14\), \(x = \frac{118.736 \pm \sqrt{1038.14}}{10} \approx 8.66, 15.1\)
    Y-values: \(y = 0.176 \times 8.66 + 2 \approx 3.53\), \(y = 0.176 \times 15.1 + 2 \approx 4.66\)
Result:
(8.66, 3.53) and (15.1, 4.66)

(f)
    Max height time: \(t = 1\) s (from 6b)
    Intersection: Use (8.66, 3.53) (before \(t = 1\))
    Ball time: \(t_{\text{ball}} = \frac{8.66 – 5}{8} = 0.4575\) s
    Arrow distance: \(\sqrt{8.66^2 + (3.53 – 2)^2} \approx 8.79\) m
    Arrow time: \(\frac{8.79}{60} \approx 0.1465\) s
    Release time: \(t_{\text{release}} = 0.4575 – 0.1465 \approx 0.311\) s
Result:
0.311 s

IB Mathematics AHL 3.12 Vector applications to kinematics-AI HL Paper 2- Exam Style Questions- New Syllabus

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