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Question 6 (12 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 6 - 2002 - Paper 1

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Question 6 (12 marks) Use a SEPARATE writing booklet. (a) An angler casts a fishing line so that the sinker is projected with a speed V m s\text{-1} from a point 5 ... show full transcript

Worked Solution & Example Answer:Question 6 (12 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 6 - 2002 - Paper 1

Step 1

i) Let (x,y) be the position of the sinker at time t seconds after the cast, and before the sinker hits the water.

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Answer

The horizontal motion is given by the equation for displacement: x=Vtcos⁡θx = V t \cos \theta To express y in terms of x, we can use the equations of motion. The vertical displacement from launch to the position just before it hits the water is derived from: y=Vtsin⁡θ−12gt2y = V t \sin \theta - \frac{1}{2} g t^2 With g being the acceleration due to gravity (10 m/s²), this simplifies to: y=Vtsin⁡θ−5t2+5.y = V t \sin \theta - 5t^2 + 5. This shows the relationship between the vertical position y and the parameters given.

Step 2

ii) Suppose the sinker hits the sea 60 metres away as shown in the diagram.

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Using the horizontal motion equation: x=60=Vtcos⁡θx = 60 = Vt \cos \theta Substituting \theta = \tan^{-1}\left(\frac{3}{4}\right)$, gives us: Vt=60cos⁡(θ)Vt = \frac{60}{\cos(\theta)} Now we find \sin(\theta) using: tan⁡(θ)=34⇒sin⁡(θ)=35,cos⁡(θ)=45\tan(\theta) = \frac{3}{4} \Rightarrow \sin(\theta) = \frac{3}{5}, \cos(\theta) = \frac{4}{5} Therefore, t=60V⋅45=75Vt = \frac{60}{V \cdot \frac{4}{5}} = \frac{75}{V} Now, substituting (t) back into the vertical motion equation, we can find the value of V algebraically.

Step 3

iii) For the cast described in part (ii), find the maximum height above sea level that the sinker achieved.

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To find the maximum height, we set the vertical velocity to zero: 0=Vsin⁡θ−gt0 = V \sin \theta - gt Solving for t at maximum height, t=Vsin⁡θgt = \frac{V \sin \theta}{g} Substituting back into the y-position function: ymax=Vtsin⁡θ−5t2+5y_{max} = Vt \sin \theta - 5t^2 + 5 We will find our maximum height using this t value and substituting the corresponding V and sin(θ) values.

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