David is in a life raft and Anna is in a cabin cruiser searching for him - HSC - SSCE Mathematics Extension 1 - Question 7 - 2003 - Paper 1

The maximum height is reached when the vertical component of the velocity becomes zero.

1. The vertical component of velocity is given by: $v_{y} = v \sin \alpha$

2. Using the equation of motion: $v_{y}^{2} = u_{y}^{2} - 2g h$ At maximum height, $v_y = 0$, thus: $0 = (v \sin \alpha)^{2} - 2gh$

3. Rearranging gives: $h = \frac{v² \sin² \alpha}{2g}$

1. The time of flight until it returns to the initial height is derived from: $y = v \sin \alpha t - \frac{1}{2} g t^{2}$ Setting y = 0 to find when it returns to height: $0 = v \sin \alpha t - \frac{1}{2} g t^{2}$

2. Factorizing gives: $t (v \sin \alpha - \frac{1}{2} g t) = 0$ Solving gives: $t = \frac{2v \sin \alpha}{g}$ (time to return to height)

3. Substitute t into x’s equation: $x = v \cos \alpha t = v \cos \alpha \left( \frac{2v \sin \alpha}{g} \right) = \frac{2v² \sin \alpha \cos \alpha}{g}$ Thus, using the identity $\sin 2α = 2 \sin α \cos α$ leads us to: $x = \frac{v²}{8} \sin 2α.$

1. For maximum separation when the ball is caught, we can relate the maximum height to the horizontal distance:

   • Using the properties of the parabolic motion and the maximum height.
   • Compute maximum height at which both can catch:
   • Generally, the separation will be maximized for the angle to catch.

2. Using similar calculations we derived earlier for height and distances gives:

   • When $v² \geq 4g(H - S)$, the distance is proportional to the parameters which can be stated as: $d = 4x (H - S).$

3. Conversely, when $v² < 4g(H - S)$, use direct relationship: $d = \frac{v²}{g}$ allowing for recovery above the catchable height.