Irodov 1.11: Same Fall Erases the Vertical Gap – Amit & Deepali Sinha Foundation

Problem Irodov 1.11

Two particles start from the same point at the same instant and fall freely under a uniform gravitational acceleration $g$ . At the start each moves horizontally, one with speed $v_1=3.0$ m/s and the other with speed $v_2=4.0$ m/s, the two directions opposite. Find the distance between the particles at the instant their velocity vectors are mutually perpendicular.

Both particles leave the common launch point O at the same instant and fall freely. Particle 1 (gold) starts horizontally at v_1, particle 2 (blue) horizontally at v_2 the opposite way; the arrow lengths carry the given 3:4 ratio. — Both particles leave the common launch point $O$ at the same instant and fall freely. Particle 1 (gold) starts horizontally at $v_1$ , particle 2 (blue) horizontally at $v_2$ the opposite way; the arrow lengths carry the given $3:4$ ratio.

Given

$v_1=3.0$ m/s
$v_2=4.0$ m/s
$g=9.8$ m/s $^2$

Find

the distance

l

between the particles when

\vect v_1\perp\vect v_2

Solution

Step 1Velocity components of each particle

Take $x$ horizontal, along particle 1’s initial velocity, and $y$ vertically down, along $\vect g$ . Both particles start at the origin with purely horizontal velocity and share the same constant downward acceleration $g$ , so their velocity components at time $t$ are

\begin{align*}\vect v_1(t) &= (v_1,\, gt) \\ \vect v_2(t) &= (-v_2,\, gt)\end{align*}

the $y$ components identical because both start from rest vertically and fall under the same $g$ .

Step 2The instant the velocities are perpendicular

Perpendicularity means the dot product of the two velocity vectors vanishes. Setting it to zero and keeping the positive root,

\vect v_1(t)\cdot\vect v_2(t) = -v_1v_2 + g^2t^2 = 0 \;\Longrightarrow\; t^2 = \frac{v_1v_2}{g^2}

t = \dfrac{\sqrt{v_1v_2}}{g}

(1)

The same pair of paths with the instant marked on each: by then each velocity has tilted away from the horizontal, and the two arrows drawn there meet at a right angle. Both marks sit at the same height. — The same pair of paths with the instant (1) marked on each: by then each velocity has tilted away from the horizontal, and the two arrows drawn there meet at a right angle. Both marks sit at the same height.

Step 3Positions, and why the vertical gap is always zero

Both particles start at rest vertically and share the same $g$ , so their vertical displacements are equal at every time; only the horizontal displacements differ, growing linearly since each horizontal velocity is constant.

\begin{align*}x_1(t) &= v_1 t, & y_1(t) &= \frac12 gt^2 \\ x_2(t) &= -v_2 t, & y_2(t) &= \frac12 gt^2\end{align*}

Since $y_1(t)=y_2(t)$ for every $t$ , the separation is purely horizontal at every instant:

l(t) = x_1(t)-x_2(t) = (v_1+v_2)t.

Step 4Evaluate the separation at the perpendicular instant

Substituting the time from (1) into the separation of Step 3 gives the algebraic result, which is then evaluated numerically.

l = (v_1+v_2)\,t

With the data,

l = \dfrac{(3.0+4.0)\sqrt{3.0\times4.0}}{9.8}\text{ m}

l = \dfrac{(v_1+v_2)\sqrt{v_1v_2}}{g} = 2.5\text{ m}

(2)

Check

Dimensionally, $[v][\sqrt{v^2}]/[g] = (\text{m/s})(\text{m/s})/(\text{m/s}^2) = \text{m}$ , a length, as required. In the equal-speed limit $v_1=v_2=v$ the perpendicularity time is $t=v/g$ , exactly the time for the vertical speed to reach the (common) horizontal speed, matching the simpler symmetric case where the velocity vector has just rotated $45^\circ$ from horizontal on each side.

Remark

The whole problem reduces to a one dimensional question in disguise: because both particles share one $g$ and both start with zero vertical velocity, their vertical motions are identical clocks, and the two dimensional separation collapses to the horizontal separation alone. That second hypothesis is doing real work and is not automatic: two particles launched obliquely with different vertical speed components (as in a problem like 1.10’s pair of projectiles) drift apart vertically too, and their separation then carries the full initial-velocity difference, not just its horizontal part.