NBPhO 2025

Overview

Both parts reduce to the same trick: eliminate gravity by going to a frame in which it does not appear.

In part (i) the drone is uniformly accelerating, so we pass to the drone’s rest frame. There gravity gains a horizontal pseudo-component, becoming an effective gravity $\vec{g}_{\mathrm{eff}}$ tilted at $45^\circ$. The target is now stationary, and — by a happy coincidence of the geometry — sits directly above the launcher along $\vec{g}_{\mathrm{eff}}$. The minimum-speed problem collapses to “throw straight up to height $H$”, which any student can solve from energy conservation.

In part (ii) the stone and the ball are both in free fall, so they share the same acceleration. Their relative motion is therefore uniform, and gravity drops out of the difference $\vec{r}_b(t)-\vec{r}_s(t)$. In the lab frame this manifests as the simple algebraic relation

$$\vec{u} \;=\; \vec{v} \,+\, \vec{D}/\tau, \qquad \vec{D}=\vec{S}-\vec{B}.$$

Minimising $|\vec{u}|$ over the intercept time $\tau$ then yields the geometric condition $\vec{u}_{\min} \perp \vec{D}$ — the boy throws perpendicular to the line from himself to the stone’s release point, not to the collision point $C$.

The two parts complement each other: part (i) shows how to remove a real acceleration by changing reference frame; part (ii) shows how to remove gravity from a problem where two objects share it.

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Part (i) — Minimum throwing speed against an accelerating drone

The drone starts at the origin at rest and accelerates with $\vec{a}_d = g\hat{x}$. Gravity is $-g\hat{y}$. The ball is thrown from $(-h,-h)$ with initial velocity $\vec{w}=(w_x,w_y)$.

Direct approach — components and a clean symmetry

Drone position: $\vec{r}_d(t) = \tfrac{1}{2}g t^2 \hat{x}$.

Ball position: $\vec{r}_b(t) = (-h+w_x t,\; -h+w_y t-\tfrac{1}{2}g t^2)$.

For collision at time $t$, both components match:

$$\tfrac{1}{2}g t^2 \;=\; -h + w_x t,$$ $$0 \;=\; -h + w_y t - \tfrac{1}{2}g t^2.$$

Adding the two equations gives $w_x t = w_y t$, i.e.

$$\boxed{w_x \;=\; w_y \quad \text{for every solution } t.}$$

This is not an extra constraint chosen by minimisation — it is a symmetry of the problem: the launch point $(-h,-h)$ lies on the diagonal $y=x$, and the relative acceleration of ball with respect to drone is symmetric under $x\leftrightarrow y$, so any intercept solution must respect that symmetry. Mistaking it for a free choice on top of the optimisation leads to over-counting unknowns.

Solving the first equation for $w_x$,

$$w_x \;=\; \frac{h}{t} + \frac{g t}{2}.$$

The squared launch speed is $w^2 = w_x^2 + w_y^2 = 2 w_x^2$, so

$$w^{2}(t) \;=\; 2\!\left(\frac{h}{t} + \frac{g t}{2}\right)^{\!2}.$$

Minimising over the intercept time

By AM–GM, $\dfrac{h}{t} + \dfrac{g t}{2} \ge 2\sqrt{\dfrac{h}{t}\cdot\dfrac{g t}{2}} = \sqrt{2gh}$, with equality at $\dfrac{h}{t}=\dfrac{g t}{2}$, i.e.

$$t_\ast \;=\; \sqrt{\tfrac{2h}{g}}.$$

Therefore

$$w^{2}_{\min} \;=\; 2(\sqrt{2gh})^{2} \;=\; 4 g h \quad\Longrightarrow\quad \boxed{\,w_{\min} \;=\; 2\sqrt{g h}\,}.$$

The optimal launch direction has $w_x = w_y = \sqrt{2 g h}$, i.e. $45^\circ$ above the horizontal, exactly bisecting the diagonal from launch point to drone.

Why the answer comes out so clean — the accelerating-frame view

The same answer drops out of a one-line argument. Pass to the drone’s (non-inertial) rest frame. There the drone sits at the origin forever, and every body is augmented by a pseudo-acceleration $-\vec{a}_d = -g\hat{x}$. The total “gravity” felt in this frame is

$$\vec{g}_{\mathrm{eff}} \;=\; -g\hat{y} \,-\, g\hat{x}, \qquad |\vec{g}_{\mathrm{eff}}| \;=\; g\sqrt{2},$$

pointing along $-(\hat{x}+\hat{y})/\sqrt{2}$ — that is, exactly along the line from the drone to the ball’s launch point $(-h,-h)$. So in this frame the launcher sits a distance $H = h\sqrt{2}$ “directly below” a stationary target, and the ball is thrown vertically upward against gravity $g\sqrt{2}$.

For a vertical throw to height $H$ in gravity $g_{\mathrm{eff}}$, energy conservation gives $w_{\min}^{2} = 2 g_{\mathrm{eff}} H$:

$$w_{\min}^{2} \;=\; 2 \cdot g\sqrt{2}\cdot h\sqrt{2} \;=\; 4gh, \qquad w_{\min} = 2\sqrt{gh}.\;\;\checkmark$$

Since the drone is at rest at $t=0$, the launch velocity transforms trivially between lab frame and drone frame, so this is also the lab-frame launch speed.

Educational remark. That the launcher–to–target line is anti-parallel to $\vec{g}_{\mathrm{eff}}$ is what makes part (i) so clean. Had the launch point been, say, $(-h,-2h)$, the target would no longer sit directly above, and one would have to solve the standard “minimum speed to reach a fixed point at $(\Delta x,\Delta y)$” problem, whose answer is $w^{2}_{\min}=g(L+\Delta y)$ with $L=\sqrt{\Delta x^{2}+\Delta y^{2}}$. The diagonal launch point $(-h,-h)$ is precisely the special configuration that lets the answer shrink to $w^{2}_{\min}=g_{\mathrm{eff}} L$ doubled.

Consistency check (dimensions and limits). $\sqrt{gh}$ has units of m s$^{-1}$. As $h \to 0$, $w_{\min}\to 0$: trivially correct. As $h\to\infty$, $w_{\min}\to\infty$: also correct. Sending the drone’s acceleration to zero (so the drone stays at the origin) would give the standard fixed-target answer $w^{2}_{\min}=g(L+\Delta y) = g(h\sqrt{2}+h)$, which is larger than $4gh$ for any $h$ — as it must be, since the drone helpfully runs away from where it would otherwise have been hit, allowing a more efficient parabolic interception.

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Part (ii) — Minimum ball speed for a moving stone target

A stone is thrown from $S$ with velocity $\vec{v}$ (magnitude $v$, direction unknown). Simultaneously, a ball is thrown from $B$ with velocity $\vec{u}$. They collide at the point $C$ shown in the figure.

I place the origin at $B$, with $+x$ horizontal toward $C$ and $+y$ vertically upward. Reading the figure (next subsection) I will denote

$$\vec{S} \;=\; (-\ell,\;h), \qquad \vec{C} \;=\; (C_x,\;C_y), \qquad \vec{D} \;\equiv\; \vec{S}-\vec{B} \;=\; (-\ell,\;h),$$

with $\ell, h, C_x, C_y$ all positive.

Why gravity disappears from the relative motion

Both projectiles undergo the same acceleration $-g\hat{y}$, so the difference of their position vectors evolves linearly:

$$\vec{r}_b(t) - \vec{r}_s(t) \;=\; (\vec{B}-\vec{S}) \,+\, (\vec{u}-\vec{v})\,t.$$

The $-\tfrac{1}{2}g\hat{y}t^{2}$ terms cancel. For collision at time $\tau$, this difference must vanish, hence

$$\boxed{\,\vec{u} \;=\; \vec{v} \,+\, \vec{D}/\tau\,}, \qquad \vec{D}=\vec{S}-\vec{B}. \tag{$\star$}$$

Equivalently: in a frame falling freely with acceleration $-g\hat{y}$, both stone and ball move at constant velocity, and the ball must travel from $\vec{B}$ to $\vec{S}$ at the relative velocity $\vec{u}-\vec{v}$ along the straight line $\vec{D}$.

The minimum-speed condition

Equation ($\star$) says the locus of admissible ball velocities is a straight line in velocity space: $\vec{u} = \vec{v} + s\hat{D}$, parametrised by $s = |\vec{D}|/\tau \in (0,\infty)$. The minimum-magnitude point on a line is its foot of perpendicular from the origin, so

$$\vec{u}_{\min} \;\perp\; \hat{D},$$

and

$$u^{2}_{\min} \;=\; v^{2} \,-\, \frac{(\vec{v}\cdot\vec{D})^{2}}{D^{2}} \;=\; v_{\perp}^{2},$$

where $v_{\perp}$ is the component of the stone’s velocity perpendicular to the line $BS$.

Two corollaries:

1. The boy throws perpendicular to $\vec{BS}$, not to $\vec{BC}$. The chord $BC$ from boy to collision is a segment of a parabola, while $\vec{BS}$ is a straight line; the launch direction is steeper than the chord. This is a notorious trap in problems of this type.

2. The optimal $\vec{u}$ is the projection of $\vec{v}$ onto the direction perpendicular to $\vec{D}$. Equivalently, $\vec{v}-\vec{u}_{\min}$ is parallel to $\vec{D}$ — when projected onto the line $BS$, the stone and ball move along $BS$ at the same rate.

Closing the system: solving for $\tau$

So far $\vec{v}$ is unknown. Two more equations come from the stone reaching $C$:

$$\vec{v}\,\tau \;=\; (\vec{C}-\vec{S}) \,+\, \tfrac{1}{2}g\hat{y}\tau^{2}. \tag{1}$$

The minimum-speed condition is $\vec{u}\cdot\vec{D}=0$. Substituting ($\star$):

$$0 \;=\; \vec{u}\cdot\vec{D} \;=\; \vec{v}\cdot\vec{D} \,+\, \frac{D^{2}}{\tau}. \tag{2}$$

Dotting (1) with $\vec{D}$ and using (2) to eliminate $\vec{v}\cdot\vec{D}$:

$$\tau\,\vec{v}\cdot\vec{D} \;=\; (\vec{C}-\vec{S})\cdot\vec{D} \,+\, \tfrac{1}{2}g\,D_y\,\tau^{2},$$ $$-D^{2} \;=\; (\vec{C}-\vec{S})\cdot\vec{D} \,+\, \tfrac{1}{2}g\,D_y\,\tau^{2}.$$

Combining $-D^{2} = -\vec{D}\cdot\vec{D}$ with $-(\vec{C}-\vec{S})\cdot\vec{D}$ on the right: $-\vec{D}\cdot[\vec{D}+(\vec{C}-\vec{S})] = -\vec{D}\cdot(\vec{C}-\vec{B}) = -\vec{D}\cdot\vec{E}$, where $\vec{E}=\vec{C}-\vec{B}$. Hence

$$\boxed{\,\tau^{2} \;=\; \frac{-2\,\vec{D}\cdot\vec{E}}{g\,D_y}.\,} \tag{3}$$

In components, with $\vec{D}=(-\ell,h)$ and $\vec{E}=(C_x,C_y)$,

$$\vec{D}\cdot\vec{E} \;=\; -\ell C_x + h C_y, \qquad D_y = h,$$

so

$$\tau^{2} \;=\; \frac{2(\ell C_x - h C_y)}{g h}. \tag{3$'$}$$

A positive $\tau^{2}$ requires $\ell C_x > h C_y$, i.e. $\vec{D}$ and $\vec{E}$ form an obtuse angle. Geometrically: the boy must stand “in the way” of the stone, with $S$ and $C$ on opposite sides of him along the BS-direction. (If $\ell C_x < h C_y$, the optimum $\tau$ is negative — meaning no positive-time intercept is optimal in this branch, and the boy simply cannot do this collision with a ball thrown in the same instant.)

Once $\tau$ is known, $\vec{v}$ follows from (1), and $\vec{u}$ from ($\star$).

Reading the figure

The figure shows points $S$, $B$, $C$ and a 10 m horizontal scale bar. Pixel measurement on the supplied PNG (the cliff foot, ground line, and the centres of the three filled dots are unambiguous) gives, with $B$ at the origin and the y-axis pointing up:

$$\ell \;\approx\; 4.87 \,\text{m}, \qquad h \;\approx\; 3.26 \,\text{m},$$ $$C_x \;\approx\; 6.53 \,\text{m}, \qquad C_y \;\approx\; 4.06 \,\text{m}.$$

The geometric setup is therefore

                                          • C  (~6.5, 4.1)
   ___________
              | S  (-4.9, 3.3)
              |
              |
              |________________________________
                         B  (0, 0)
                         |---------- 10 m ---------|

The boy stands roughly $5\,\text{m}$ to the right of the cliff foot, on ground $\sim 3.3\,\text{m}$ below $S$. The collision $C$ is about $6.5\,\text{m}$ to the boy’s right and $\sim 4\,\text{m}$ above his head — so $C$ is higher than $S$ and well off to the right.

A reasonable error bar on each coordinate from figure reading is $\pm 0.2\,\text{m}$. As we shall see, this propagates to a few percent in $\tau$, $v$ and $u$.

Numerical evaluation

Using $g=9.8\,\text{m s}^{-2}$ and the readings above:

$$\ell C_x \;=\; 4.87\cdot 6.53 \;\approx\; 31.8 \,\text{m}^{2},\qquad h C_y \;=\; 3.26\cdot 4.06 \;\approx\; 13.2 \,\text{m}^{2},$$ $$\tau^{2} \;=\; \frac{2(31.8-13.2)}{9.8\cdot 3.26} \;=\; \frac{2\cdot 18.6}{31.95} \;\approx\; 1.16 \,\text{s}^{2}, \qquad \tau \;\approx\; 1.08 \,\text{s}.$$

Pitfall. The numerator $\ell C_x - h C_y$ is a difference of comparable quantities ($\sim 32$ and $\sim 13$). A 4% error in either term shifts $\tau^{2}$ by $\sim 7\%$ and $\tau$ by $\sim 3\text{–}4\%$. This is what limits the precision below.

Stone speed

From (1):

$$v_x \;=\; \frac{C_x - S_x}{\tau} \;=\; \frac{6.53-(-4.87)}{1.08} \;\approx\; \frac{11.40}{1.08} \;\approx\; 10.6 \,\text{m s}^{-1},$$ $$v_y \;=\; \frac{C_y - S_y}{\tau} + \tfrac{1}{2} g \tau \;=\; \frac{4.06-3.26}{1.08} + \tfrac{1}{2}(9.8)(1.08) \;\approx\; 0.74 + 5.28 \;\approx\; 6.02 \,\text{m s}^{-1}.$$ $$v \;=\; \sqrt{v_x^{2} + v_y^{2}} \;=\; \sqrt{111.7 + 36.3} \;\approx\; \sqrt{148} \;\approx\; 12.2 \,\text{m s}^{-1}.$$

The stone’s launch angle is $\arctan(v_y/v_x)\approx\arctan(0.57)\approx 30^\circ$ above horizontal — a low, fast throw aimed up and to the right. (A complementary angle $\sim 60^\circ$ would also reach $C$, but only at a different $\tau$ and different $v$, and would not be the optimum the boy is exploiting.)

Ball speed (the answer)

From ($\star$) with $\vec{D}/\tau=(-4.87/1.08,\;3.26/1.08)\approx(-4.51,\;3.02)\,\text{m s}^{-1}$:

$$u_x \;=\; v_x + D_x/\tau \;\approx\; 10.6 - 4.51 \;\approx\; 6.06 \,\text{m s}^{-1},$$ $$u_y \;=\; v_y + D_y/\tau \;\approx\; 6.02 + 3.02 \;\approx\; 9.05 \,\text{m s}^{-1},$$ $$u \;=\; \sqrt{u_x^{2}+u_y^{2}} \;=\; \sqrt{36.7+81.9} \;\approx\; \sqrt{118.6} \;\approx\; 10.9 \,\text{m s}^{-1}.$$

The ball leaves $B$ at angle $\arctan(u_y/u_x)\approx 56^\circ$ above horizontal.

Independent checks

Perpendicularity, $\vec{u}\cdot\vec{D}=0$:

$$\vec{u}\cdot\vec{D} \;\approx\; 6.06\cdot(-4.87) + 9.05\cdot 3.26 \;\approx\; -29.5 + 29.5 \;\approx\; 0.\;\;\checkmark$$

Angle of $\vec{u}$ vs angle of $\vec{D}$: the line $\vec{BS}$ makes an angle $\arctan(h/\ell)\approx\arctan(3.26/4.87)\approx 33.8^\circ$ above horizontal (measured into the upper-left quadrant). The perpendicular direction is $90^\circ - 33.8^\circ = 56.2^\circ$ above horizontal — exactly the launch angle we found for $\vec{u}$. $\checkmark$

The shortcut formula $u_{\min}=v_{\perp}$: with $\hat{D}=(-\ell,h)/|\vec{D}|=(-4.87,\,3.26)/5.86 \approx (-0.831,\,0.557)$,

$$v_{\parallel} \;=\; \vec{v}\cdot\hat{D} \;\approx\; 10.6(-0.831)+6.02(0.557) \;\approx\; -8.81 + 3.35 \;\approx\; -5.46\,\text{m s}^{-1},$$ $$v_{\perp} \;=\; \sqrt{v^{2} - v_{\parallel}^{2}} \;\approx\; \sqrt{148 - 29.8} \;\approx\; \sqrt{118.2} \;\approx\; 10.9\,\text{m s}^{-1}.\;\checkmark$$

(The negative sign of $v_{\parallel}$ confirms the assumption $\vec{v}\cdot\vec{D}<0$ used in the optimisation: the stone has a component of velocity toward $B$ along the line $BS$, which is what allows $\tau>0$ at the optimum.)

Don’t use the fixed-target formula here. A common mistake is to plug $C$ into the well-known minimum speed for fixed target $L$, $\Delta y$: $u^{2}_{\text{fixed}} = g(L+\Delta y)$, with $L=|\vec{C}-\vec{B}|\approx 7.69\,\text{m}$ and $\Delta y = C_y\approx 4.06\,\text{m}$, giving $u_{\text{fixed}} \approx \sqrt{9.8\cdot 11.75}\approx 10.7\,\text{m s}^{-1}$. This is numerically close to our answer, but only by accident — it would be the answer if the stone hovered at $C$, whereas in reality the stone is moving when intercepted, and the correct answer $\sqrt{v^{2}-v_{\parallel}^{2}}$ depends on the stone’s velocity, not just the geometry of $C$.

Uncertainty

A figure-reading error of $\delta\sim 0.2\,\text{m}$ in $C_x$, $C_y$, $\ell$, $h$ propagates to

$$\frac{\delta(\tau^{2})}{\tau^{2}} \;\sim\; \frac{\sqrt{(\ell\,\delta C_x)^{2} + (C_x\,\delta\ell)^{2} + (h\,\delta C_y)^{2} + (C_y\,\delta h)^{2}}}{\ell C_x - h C_y}\;\sim\;\frac{1.7\,\text{m}^{2}}{18.6\,\text{m}^{2}}\sim 9\%.$$

So $\tau$ is good to $\sim 4\text{–}5\%$ and the speeds $u,v$ to a similar relative tolerance. Three significant figures in the final answer would over-state the precision; two significant figures is the honest quote:

$$\boxed{\,v \;\approx\; 12 \,\text{m s}^{-1}, \qquad u \;\approx\; 11 \,\text{m s}^{-1}.\,}$$

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Summary of results

Part	Quantity	Symbolic answer	Numerical value
(i)	Minimum ball speed to hit drone	$w_{\min} = 2\sqrt{gh}$	(depends on $h$)
(ii)	Stone speed	$v = \sqrt{v_x^{2}+v_y^{2}}$ from eq. (1)	$v \approx 12 \,\text{m s}^{-1}$
(ii)	Minimum ball speed	$u = \sqrt{v^{2} - (\vec{v}\cdot\hat{D})^{2}} = v_{\perp}$	$u \approx 11 \,\text{m s}^{-1}$
(ii)	Time of collision	$\tau^{2} = \dfrac{-2\,\vec{D}\cdot\vec{E}}{g\,D_y}$	$\tau \approx 1.1 \,\text{s}$

The unifying lesson: in projectile-intercept problems where two bodies share the same gravitational acceleration, transform to a free-fall frame; in problems where one body has an additional acceleration, transform to that body’s rest frame and absorb the pseudo-acceleration into an effective gravity. After the transform, what is left is a kinematic problem about straight lines.