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

To find V, we set x = 60. From the horizontal motion equation:

$60 = Vt \cos(\theta)$

Given that θ = tan⁻¹(3/4), we can find cos(θ):

$\cos(\theta) = \frac{4}{5}$

Thus:

$60 = Vt \frac{4}{5}$

This gives us:

$Vt = 75\implies t = \frac{75}{V}$

Now substituting t in the equation for vertical motion:

$y = Vt \sin(\theta) - 5t^2 + 5$

We know y = 0 when it hits the water:

$0 = V \left(\frac{75}{V}\right) \cdot \frac{3}{5} - 5 \left(\frac{75}{V}\right)^2 + 5$

Solving for V leads to:

$0 = 45 - \frac{28125}{V^2} + 5$

This simplifies down to:

$\frac{28125}{V^2} = 50 \implies V^2 = 562.5 \implies V = 7.5$

To find the maximum height, we first set the time to reach maximum height. This occurs when the vertical velocity is 0:

$V \sin(\theta) - gt = 0$

Thus,

$t_{max} = \frac{V \sin(\theta)}{g} = \frac{7.5 \cdot \frac{3}{5}}{10} = 0.225s$

Substituting this back into the vertical motion equation to find the height:

$y_{max} = V \cdot 0.225 \cdot \frac{3}{5} - 5(0.225)^2 + 5$

Calculating this gives:

$y_{max} = 2.25 - 0.5625 + 5 = 6.6875m$

Thus, the maximum height above sea level is approximately 6.69m.