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

Given that $x = 60$, we use the horizontal motion equation:

$x = Vt \cos\theta.$
Replace $x$ with 60:

$60 = Vt \cos\left(\tan^{-1}\frac{3}{4}\right).$
Calculating $\cos\left(\tan^{-1}\frac{3}{4}\right)$, we find:

$\cos\left(\tan^{-1}\frac{3}{4}\right) = \frac{4}{5}.$
Thus, the equation becomes:

$60 = Vt \frac{4}{5} \implies Vt = 75,$
which leads to:

$v = \frac{75}{t}.$
To find $t$, substitute in the vertical motion equation with $y = 0$ (sea level), giving:

$0 = Vt \sin\left(\tan^{-1}\frac{3}{4}\right) - 5t^{2} + 5.$
Calculating:

$\sin\left(\tan^{-1}\frac{3}{4}\right) = \frac{3}{5},$
Substituting gives:

$0 = \frac{75}{t} \cdot \frac{3}{5} - 5t^{2} + 5.$
After some algebraic manipulation, we will solve for $V$.

To find the maximum height, we know the maximum occurs when the vertical velocity is zero:

$v_y = V \sin\theta - gt = 0.$
Solving for $t$, we find:

$t = \frac{V \sin\theta}{g} = \frac{V \cdot \frac{3}{5}}{10} = \frac{3V}{50}.$
Now we substitute $t$ back into the vertical displacement equation:

$y_{max} = y_0 + Vt \sin\theta - 5t^2$
$y_{max} = 5 + V \left(\frac{3V}{50}\right) \cdot \frac{3}{5} - 5 \left(\frac{3V}{50}\right)^2.$
This expression can be simplified to find the maximum height achieved by the sinker above sea level.