A ship moves in a straight line with constant speed v . A…

A ship moves in a straight line with constant speed $v$. A projectile is launched from the ship with speed $v_0$ (relative to the water) in the direction of the ship's motion, at an angle $\alpha$ to the horizontal. (a) How far from the ship does the projectile land in the water? (b) At what angle $\alpha_2$ must it be launched to land back on the ship? Give the trajectory in that case. (c) What is the flight time in each case? (d) In case (b), what must the ship's speed be for the projectile's maximum height to equal the distance the ship travels until the projectile lands on it? Ignore the ship's size.

Explanation of Method

A projectile fired from a moving ship.

Step 1: Flight time (c)

The vertical motion does not depend on the ship at all: $t_f = \frac{2v_0\sin\alpha}{g}$.

In case (b) this becomes

\[ t_f = \frac{2v_0\sin\alpha_2}{g} = \frac{2\sqrt{v_0^2 - v^2}}{g} \]

Step 2: (a) Distance from the ship

The projectile moves forward at $v_0\cos\alpha$, the ship at $v$. The gap between them grows at the difference: $d = (v_0\cos\alpha - v)\,t_f = \frac{v_0^2\sin 2\alpha}{g} - \frac{2vv_0\sin\alpha}{g}$.

\[d = \frac{2v_0\sin\alpha\,(v_0\cos\alpha - v)}{g}\]

(Ahead of the ship if $v_0\cos\alpha \gt v$, behind it otherwise.)

$\text{(a) } v_0\cos\alpha > v$ $v$ $d$ $\text{(b) } v_0\cos\alpha_2 = v$ $v$ $\text{lands on the ship}$

Seen from the shore (example numbers, to scale). The projectile moves forward at $v_0\cos\alpha$, the ship at $v$; the dashed ship shows where it is at the landing moment. (a) The projectile outruns the ship by $d$. (b) With $v_0\cos\alpha_2 = v$ they stay level horizontally, so it comes down on the ship.

Step 3: (b) Landing back on the ship

The gap stays zero if the horizontal speeds are equal: $v_0\cos\alpha_2 = v \;\Longrightarrow\; \alpha_2 = \arccos\frac{v}{v_0}$.

With the launch point as origin (fixed to the water), the trajectory is

\[ y = x\tan\alpha_2 - \frac{g\,x^2}{2v_0^2\cos^2\alpha_2} = x\,\frac{\sqrt{v_0^2 - v^2}}{v} - \frac{g\,x^2}{2v^2}. \]

Seen from the ship, the projectile simply goes straight up and falls straight back down.

Step 4: (d) Height equal to the ship's travel

\[ H = \frac{v_0^2\sin^2\alpha_2}{2g} = \frac{v_0^2 - v^2}{2g}, \]

\[ s_{\text{ship}} = v\,t_f = \frac{2v\sqrt{v_0^2 - v^2}}{g}. \]

\[ H = s_{\text{ship}} \;\Longrightarrow\; \sqrt{v_0^2 - v^2} = 4v \;\Longrightarrow\; v_0^2 = 17\,v^2. \]

\[v = \frac{v_0}{\sqrt{17}} \approx 0.243\,v_0 \qquad (\tan\alpha_2 = 4,\ \alpha_2 \approx 76^\circ)\]

Result

\[\boxed{\,t_f = \frac{2v_0\sin\alpha}{g}\,}\]

\[\boxed{\,d = \frac{2v_0\sin\alpha\,(v_0\cos\alpha - v)}{g}\,}\]

\[\boxed{\,v_0\cos\alpha_2 = v \;\Longrightarrow\; \alpha_2 = \arccos\frac{v}{v_0}\,}\]

\[\boxed{\,v = \frac{v_0}{\sqrt{17}} \approx 0.243\,v_0 \qquad (\tan\alpha_2 = 4,\ \alpha_2 \approx 76^\circ)\,}\]

Remark

About the original solution. In (a) it added the ship's travel $v\,t_f$ to the range; that gives the landing point measured from where the ship was at launch, not the distance from the ship. Since the ship also moves, its travel must be subtracted. Its (b) and (c) are correct, and its (d) result $v = \tfrac{v_0\sin\alpha}{4}$ is equivalent to $v = v_0/\sqrt{17}$ once $\sin\alpha_2 = \sqrt{1 - v^2/v_0^2}$ is used. (This is the same angle $\tan\alpha = 4$ as in Gravitational Kinematics 032.)