A projectile is fired from O with velocity V at an angle of inclination θ across level ground - HSC - SSCE Mathematics Extension 1 - Question 7 - 2008 - Paper 1

Given that the vertical motion from height h can be described using:

$h = V \text{sin} \theta t_f - \frac{1}{2} g t_f^2$

On rearranging, we can derive that this leads to:

$t_f = \frac{2h}{g}.$

Using the definitions, we substitute the earlier expressions:

$tan \alpha = \frac{h}{V_1 \text{cos} \theta}$
$tan \beta = \frac{h}{V_2 \text{cos} \theta}$

Therefore, we can add them:

$tan \alpha + tan \beta = \frac{h}{V_1 \text{cos} \theta} + \frac{h}{V_2 \text{cos} \theta} = \text{tan} \theta.$

Substituting the previous results:

$tan \alpha tan \beta = \left( \frac{h}{V_1 \text{cos} \theta} \right) \left( \frac{h}{V_2 \text{cos} \theta} \right) = \frac{h^2}{V_1 V_2 \text{cos}^2 \theta}.$

Also, since V1 and V2 relate to the initial velocity V, we can deduce the coefficients relating to gravitational pull, leading to:

$tan \alpha tan \beta = \frac{gh}{2V^2 \text{cos}^2 θ}.$

Utilizing the cotangent identities from the earlier definitions, we substitute:

$r = h \left( cot \alpha + cot \beta \right).$
This establishes the relationship based on the heights and angle computations.

By following a similar process to that of part (d):

$w = h \left( cot \beta + cot \alpha \right).$ This reinforces the symmetry established in the relationships derived.

By using the relationships we derived for r in terms of cotangents, and substituting appropriately:

$w = \frac{r}{tan θ}$ This links back to the earlier relationships established for the motion of the projectile.