SSCE HSC Mathematics Extension 1 Definite integrals and substitution: An angler casts a fishing line s

An angler casts a fishing line so that the sinker is projected with a speed V m s$^{-1}$ from a point 5 metres above a flat sea - HSC - SSCE Mathematics Extension 1 - Question 6 - 2002 - Paper 1

Question 6

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An angler casts a fishing line so that the sinker is projected with a speed V m s$^{-1}$ from a point 5 metres above a flat sea. The angle of projection to the horiz... show full transcript

Worked Solution & Example Answer:An angler casts a fishing line so that the sinker is projected with a speed V m s$^{-1}$ from a point 5 metres above a flat sea - HSC - SSCE Mathematics Extension 1 - Question 6 - 2002 - Paper 1

Step 1

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

To derive the equation for $y(t)$ , we start with the known equations of motion for the sinker:

The horizontal position is given by:
$x = Vt \, ext{cos} \, heta$
Since the vertical motion is under the influence of gravity, we have: $y(t) = h_0 + Vt \, ext{sin} \, heta - rac{1}{2} g t^2$
where $h_0 = 5$ m is the initial height above sea level and $g = 10$ m/s $^2$ .
Incorporating this, we get: $y = 5 + Vt \, ext{sin} \, heta - 5t^2$
This simplifies to the required form, showing the relationship between $y$ and $t$ .

Step 2

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

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Answer

To find the value of $V$ when the sinker hits the sea:

From the horizontal motion, we know: $x = 60 = Vt \, ext{cos} \, \theta$
Given that $heta = \tan^{-1} \frac{3}{4}$ , we can calculate:

ightarrow \text{cos} \theta = \frac{4}{5} , ext{and} , \text{sin} \theta = \frac{3}{5} $$

Replacing in our earlier equation, we get: $60 = Vt \frac{4}{5}$
Rearranging gives: $V = \frac{60 \cdot 5}{4t} = \frac{75}{t}$
From the vertical motion when $y = 0$ : $0 = 5 + Vt \frac{3}{5} - 5t^2$ Setting $y = 0$ leads to: $0 = 5 + \frac{75t}{t} \cdot \frac{3}{5} - 5t^2$
This leads to a quadratic equation which can be solved for $t$ , then used to find $V$ .

Step 3

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

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Answer

The maximum height occurs when the vertical velocity is zero:

Vertical velocity is given by: $v_y = V \sin(\theta) - g t$
Setting $v_y = 0$ , we obtain: $g t = V \sin(\theta)$ So, $t = \frac{V \cdot \sin(\theta)}{g}$
Substitute this time into the height equation: $y_{max} = 5 + V \sin(\theta) t - 5t^2$
Finally plug the obtained value of $t$ into this equation to compute $y_{max}$ and identify the height.

An angler casts a fishing line so that the sinker is projected with a speed V m s$^{-1}$ from a point 5 metres above a flat sea - HSC - SSCE Mathematics Extension 1 - Question 6 - 2002 - Paper 1

Worked Solution & Example Answer:An angler casts a fishing line so that the sinker is projected with a speed V m s$^{-1}$ from a point 5 metres above a flat sea - HSC - SSCE Mathematics Extension 1 - Question 6 - 2002 - Paper 1

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

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

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

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