The graphs of the functions $y = kx^n$ and $y = ext{log}_e x$ have a common tangent at $x = a$, as shown in the diagram - HSC - SSCE Mathematics Extension 1 - Question 7 - 2007 - Paper 1

From the equation $a^n = \frac{1}{nk}$, we can express $k$ as: $k = \frac{1}{na^n}.$ To eliminate $a$, we substitute $a$ from the first equation into this expression. If needed, we can use another expression that defines the relationship between $k$, $n$, and the variable $a$ more fully.

Substituting the equations of motion into the relationship:

• From $y = 14t \sin \theta - 4.9t^2$ and $x = 14t \cos \theta$, we can eliminate $t$:
• Setting $t = \frac{x}{14 \cos \theta}$ into $y$ gives:

$y = 14 \frac{x}{14 \cos \theta} \sin \theta - 4.9 \left( \frac{x}{14 \cos \theta} \right)^2$ Simplifying gives the required equation $y = mx - \left( \frac{1 + m^2}{40} \right)x^2$.

To determine when the paintball hits the barrier, we must set $y$ equal to the barrier height ($h$): Substituting our equation of trajectory into: $h = mx - \left( \frac{1 + m^2}{40} \right)x^2$ After rearranging and solving for $m$, we arrive at: $m = 2 \pm \sqrt{3 - 0.4h}.$

Given the first interval $2.8 \leq x \leq 3.2$, we can analyze the conditions where the paintball hits the barrier:

• We need to determine if there are other solutions for $x$ by checking the trajectory equation and the height of the barrier. By factoring or applying the quadratic formula, we can find the range that satisfies the criteria, leading to the second interval.

Setting $y = 0$ in $y = mx - \left( \frac{40}{1 + m^2} \right)$ will give us the equations to calculate the corresponding ground values for $x$ from both the intervals previously determined. The specific widths will depend on the boundaries set by the height at which it travels through the hole, ensuring we derive a consistent range upon substitution.