1. **Problem statement:** We have a falling object with velocity $v(t)$ under constant gravitational acceleration $g=10$ m/s², described by the differential equation $v' = -10$. The initial velocity is $v(0) = 1000$ m/s upwards. 2. **Solve for velocity $v(t)$:** The differential equation is $$v' = \frac{dv}{dt} = -10.$$ Integrate both sides with respect to $t$: $$\int dv = \int -10 \, dt$$ $$v = -10t + C$$ Use the initial condition $v(0) = 1000$: $$1000 = -10 \cdot 0 + C \implies C = 1000$$ So, $$v(t) = 1000 - 10t.$$ 3. **Find height $s(t)$ as a function of time:** Velocity is the derivative of height: $$v = \frac{ds}{dt} = 1000 - 10t.$$ Integrate velocity to get height: $$s(t) = \int (1000 - 10t) dt = 1000t - 5t^2 + D.$$ Assuming the bullet is fired from ground level, $s(0) = 0$: $$0 = 1000 \cdot 0 - 5 \cdot 0^2 + D \implies D = 0.$$ Thus, $$s(t) = 1000t - 5t^2.$$ 4. **Find the time when the bullet hits the ground:** Set $s(t) = 0$: $$1000t - 5t^2 = 0$$ Factor out $t$: $$t(1000 - 5t) = 0$$ Solutions: $$t = 0 \quad \text{(launch time)}$$ $$1000 - 5t = 0 \implies t = \frac{1000}{5} = 200 \, \text{seconds}.$$ 5. **Find velocity when bullet lands:** Evaluate $v(t)$ at $t=200$: $$v(200) = 1000 - 10 \cdot 200 = 1000 - 2000 = -1000 \, \text{m/s}.$$ The negative sign indicates downward velocity. 6. **Differential equation with air resistance proportional to velocity squared:** Air resistance force opposes motion and is proportional to $v^2$ with sign depending on direction. The differential equation is: $$v' = -10 - k v |v|,$$ where $k > 0$ is a constant of proportionality. This ensures resistance always opposes velocity whether going up or down. **Final answers:** - Velocity function: $$v(t) = 1000 - 10t$$ - Height function: $$s(t) = 1000t - 5t^2$$ - Velocity on landing: $$-1000$$ meters per second - Air resistance differential equation: $$v' = -10 - k v |v|$$