Solve $$\frac{3x}{x-2} \leq 1$$ 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

From the problem:

• The horizontal distance is given as $x = 4000 m$,
• The relationship between $x$ and $t$ is given by: $x = Vt$ So, $V = \frac{x}{t}$.

The vertical distance can also be expressed: From the equation for $y$, $y = -5t^2.$
At the moment before impact, when $y = 0$, we set: $0 = -5t^2 \Rightarrow t = \sqrt{0} = 0 \\ \Rightarrow \text{not valid, we look for the time when it hits the ground. Note for 45 degrees: } \tan(45^{\circ}) = 1 \Rightarrow \frac{y}{x} \Rightarrow y = x.$
Given that the bomb hits the ground $4000 m$ horizontally away at $45^{\circ}$, the equations of projectile motion apply. We need to find the time taken to drop this vertical distance from the height $h$: Substituting back, we can find the exact numbers for motion to calculate $V$. Resolving the values, we consider:

• For $y$ to be similar in coordinate system, it means solving specifically for time taken at the respective coordinates which solves within given conditions, thus $x = 4000$ meets conditions at horizontal speed.

We can solve this first-order differential equation using separation of variables. $\frac{dx}{x} = -4dt.$ Integrating both sides: $\ln|x| = -4t + C,$ where $C$ is the constant of integration. Exponentiating both sides: $x = e^{-4t + C} = ke^{-4t},$ where $k = e^C$ is a constant.

To find $k$, we use the initial conditions provided at time $t=0$. Given $x = 3$, Therefore: $3 = ke^{0}, \Rightarrow k = 3.$ Hence, we have: $x = 3e^{-4t}.$ Considering further condition of $x = -6/3$, solving the constant conditions again provides:
In similarly derived cases for harmonic system of $x$ gives out final conditions for specific resonation values at completing oscillation motion.