Solution 22
1. Drag Model Interpretation
The problem specifies a speed reduction rate $k = 0.1 \text{ s}^{-1}$ (implied from units and context). This corresponds to a linear drag model where the horizontal acceleration is $a_x = -k v_x$.
The problem specifies a speed reduction rate $k = 0.1 \text{ s}^{-1}$ (implied from units and context). This corresponds to a linear drag model where the horizontal acceleration is $a_x = -k v_x$.
2. Horizontal Range Calculation
Solving the differential equation for horizontal velocity: $$ v_x(t) = u_x e^{-kt} $$ The horizontal position $x(t)$ is: $$ x(t) = \int_0^t v_x dt = \frac{u_x}{k} (1 – e^{-kt}) $$ Given that the cliff is “very high”, the stone will be in the air for a long time ($t \to \infty$). The maximum horizontal distance the stone can travel is the limiting value of $x(t)$: $$ x_{\text{max}} = \lim_{t \to \infty} \frac{u_x}{k} (1 – e^{-kt}) = \frac{u_x}{k} $$
Solving the differential equation for horizontal velocity: $$ v_x(t) = u_x e^{-kt} $$ The horizontal position $x(t)$ is: $$ x(t) = \int_0^t v_x dt = \frac{u_x}{k} (1 – e^{-kt}) $$ Given that the cliff is “very high”, the stone will be in the air for a long time ($t \to \infty$). The maximum horizontal distance the stone can travel is the limiting value of $x(t)$: $$ x_{\text{max}} = \lim_{t \to \infty} \frac{u_x}{k} (1 – e^{-kt}) = \frac{u_x}{k} $$
3. Calculation
Initial horizontal velocity $u_x = u \cos\theta$. Given: $u = 40$ m/s, $\theta = 60^\circ$, $k = 0.1$ s$^{-1}$. $$ u_x = 40 \cos 60^\circ = 40(0.5) = 20 \text{ m/s} $$ $$ x_{\text{max}} = \frac{20}{0.1} = 200 \text{ m} $$
Initial horizontal velocity $u_x = u \cos\theta$. Given: $u = 40$ m/s, $\theta = 60^\circ$, $k = 0.1$ s$^{-1}$. $$ u_x = 40 \cos 60^\circ = 40(0.5) = 20 \text{ m/s} $$ $$ x_{\text{max}} = \frac{20}{0.1} = 200 \text{ m} $$
Answer:
The soldier should install the catapult at a horizontal separation of 200 m from the enemy camps.
The soldier should install the catapult at a horizontal separation of 200 m from the enemy camps.