Solve $$\frac{3x}{x - 2} \leq 1.$$ --- (b) An aircraft flying horizontally at $V \text{ m s}^{-1}$ releases a bomb that hits the ground 4000 m away, measured horizontally - HSC - SSCE Mathematics Extension 1 - Question 4 - 2001 - Paper 1

For the bomb to hit the ground at an angle of 45°, we know that the horizontal and vertical distances traveled must be equal when it reaches the ground. Given that:

• Horizontal distance $x = 4000$ m
• Vertical distance $y = -5t^2$ when it hits the ground at $t$ seconds: $y(t) = -5t^2 = 0 \Rightarrow t = \sqrt{\frac{4000}{V}}.$
  Since it hits the ground at 45°, we have: $\tan(45°) = \frac{y}{x} = 1 \Rightarrow\ 4000 = -5t^2,$
  Substituting for $t$ we find: $4000 = -5 \left(\frac{4000}{V}\right) \Rightarrow V = 2000 \text{ m s}^{-1}.$

This is a separable differential equation. We can separate and integrate: $\frac{dx}{x} = -4dt.$
Integrating both sides gives: $\ln|x| = -4t + C.$
Exponentiating: $x = e^{-4t + C} = Ce^{-4t}.$
Using the initial condition $x = 3$ when $t = 0$ gives: $3 = Ce^0 \Rightarrow C = 3.$
Thus: $x(t) = 3e^{-4t}.$ To find the specific point when $t = 0$ again, we check the value: $x(0) = 3,$ confirming our function.