Source: High School Physics Olympiad
Problem Sets:
Problem
A boat moves along a river in uniform rectilinear motion with speed $v$. A person on the boat throws an object obliquely forward (in the direction of the boat's motion) with speed $v_0$ relative to the boat, where $v_0$ makes an angle $\theta$ with the horizontal.
- Taking the Earth as the reference frame, find the trajectory equation of the object.
- Taking the boat as the reference frame, find the trajectory equation of the object.
Earth frame: $y = \dfrac{v_0\sin\theta}{v + v_0\cos\theta}\,x - \dfrac{g\,x^2}{2(v + v_0\cos\theta)^2}$; boat frame: $y = x'\tan\theta - \dfrac{g\,x'^2}{2v_0^2\cos^2\theta}$
Take the launch point as the origin, $x$ horizontal in the direction of motion, $y$ vertically upward.
Earth frame: The initial velocity is the vector sum of the boat's velocity and the relative throw velocity: horizontal component $v + v_0\cos\theta$, vertical component $v_0\sin\theta$. The motion equations are:
$$x = (v + v_0\cos\theta)t, \qquad y = v_0\sin\theta\,t - \frac{1}{2}gt^2$$Eliminating $t = \dfrac{x}{v + v_0\cos\theta}$:
$$y = \frac{v_0\sin\theta}{v + v_0\cos\theta}\,x - \frac{g}{2(v + v_0\cos\theta)^2}\,x^2$$Boat frame: The boat moves uniformly, so it is also an inertial frame and the object undergoes ordinary projectile motion with initial speed $v_0$ at angle $\theta$:
$$x' = v_0\cos\theta\,t, \qquad y = v_0\sin\theta\,t - \frac{1}{2}gt^2$$Eliminating $t$:
$$y = x'\tan\theta - \frac{g}{2v_0^2\cos^2\theta}\,x'^2$$Both trajectories are parabolas; the Earth-frame parabola is stretched horizontally by the boat's motion.