Solution 21
1. Equations of Motion
The acceleration vector is $\vec{a} = -k\vec{v} + \vec{g}$. Resolving into components ($u_x = u\cos\theta, u_y = u\sin\theta$): $$ \frac{dv_x}{dt} = -k v_x $$ $$ \frac{dv_y}{dt} = -k v_y – g $$
The acceleration vector is $\vec{a} = -k\vec{v} + \vec{g}$. Resolving into components ($u_x = u\cos\theta, u_y = u\sin\theta$): $$ \frac{dv_x}{dt} = -k v_x $$ $$ \frac{dv_y}{dt} = -k v_y – g $$
2. Horizontal Motion ($x$)
Integrating the horizontal equation: $$ \int_{u_x}^{v_x} \frac{dv}{v} = -k \int_0^t dt \implies v_x = u_x e^{-kt} $$ Integrating velocity to find position $x(t)$: $$ x(t) = \int_0^t u_x e^{-k\tau} d\tau = \frac{u_x}{k} (1 – e^{-kt}) $$
Integrating the horizontal equation: $$ \int_{u_x}^{v_x} \frac{dv}{v} = -k \int_0^t dt \implies v_x = u_x e^{-kt} $$ Integrating velocity to find position $x(t)$: $$ x(t) = \int_0^t u_x e^{-k\tau} d\tau = \frac{u_x}{k} (1 – e^{-kt}) $$
3. Vertical Motion ($y$)
Rearranging the vertical equation: $\frac{dv_y}{dt} + k v_y = -g$. Using the integrating factor $e^{kt}$: $$ \frac{d}{dt}(v_y e^{kt}) = -g e^{kt} $$ $$ v_y e^{kt} = -\frac{g}{k}e^{kt} + C $$ At $t=0, v_y = u_y \implies C = u_y + \frac{g}{k}$. Thus, $v_y = \left( u_y + \frac{g}{k} \right)e^{-kt} – \frac{g}{k}$. Integrating for position $y(t)$: $$ y(t) = \int_0^t \left[ \left( u_y + \frac{g}{k} \right)e^{-k\tau} – \frac{g}{k} \right] d\tau $$ $$ y(t) = \left( \frac{u_y + g/k}{k} \right)(1 – e^{-kt}) – \frac{g}{k}t $$
Rearranging the vertical equation: $\frac{dv_y}{dt} + k v_y = -g$. Using the integrating factor $e^{kt}$: $$ \frac{d}{dt}(v_y e^{kt}) = -g e^{kt} $$ $$ v_y e^{kt} = -\frac{g}{k}e^{kt} + C $$ At $t=0, v_y = u_y \implies C = u_y + \frac{g}{k}$. Thus, $v_y = \left( u_y + \frac{g}{k} \right)e^{-kt} – \frac{g}{k}$. Integrating for position $y(t)$: $$ y(t) = \int_0^t \left[ \left( u_y + \frac{g}{k} \right)e^{-k\tau} – \frac{g}{k} \right] d\tau $$ $$ y(t) = \left( \frac{u_y + g/k}{k} \right)(1 – e^{-kt}) – \frac{g}{k}t $$
Final Expressions:
Coordinates: $$ x = \frac{u \cos\theta}{k}(1 – e^{-kt}) $$ $$ y = \left( \frac{u \sin\theta}{k} + \frac{g}{k^2} \right)(1 – e^{-kt}) – \frac{g}{k}t $$ Terminal Velocity (as $t \to \infty$): $$ v_{term} = \frac{g}{k} \text{ (downwards)} $$
Coordinates: $$ x = \frac{u \cos\theta}{k}(1 – e^{-kt}) $$ $$ y = \left( \frac{u \sin\theta}{k} + \frac{g}{k^2} \right)(1 – e^{-kt}) – \frac{g}{k}t $$ Terminal Velocity (as $t \to \infty$): $$ v_{term} = \frac{g}{k} \text{ (downwards)} $$