Irodov 1.165: The Falling Frame Meets Relative Rest – Amit & Deepali Sinha Foundation

Problem Irodov 1.165

A stone is released from rest at height $h$ above the ground and falls freely under a constant gravitational acceleration $g$ , air resistance neglected, striking the ground with speed $v_0=\sqrt{2gh}$ relative to the Earth. Demonstrate that this same relation for $v_0$ is recovered when the fall is analyzed instead in a reference frame $K'$ that itself moves downward, relative to the Earth, at the constant velocity $v_0$ .

The setup in the Earth frame K: the stone is released from rest a height h above the ground and falls freely from there. — The setup in the Earth frame $K$ : the stone is released from rest a height $h$ above the ground and falls freely from there.

Given

$h$
$g$
K $'$ moves at constant velocity $v_0=\sqrt{2gh}$ relative to the Earth

Find

demonstrate that the same relation

v_0=\sqrt{2gh}

follows when the fall is analyzed in

K'

Solution

Step 1Transform the stone’s and the ground’s velocities into

K'

Take the downward direction as positive, with the stone released at $t=0$ from $y=0$ and the ground at $y=h$ . In $K$ the stone’s velocity is $u(t)=gt$ , and the ground, fixed in $K$ , has velocity $0$ . Since $K'$ moves at the constant velocity $v_0$ (downward) relative to $K$ , subtracting $v_0$ from each gives the same quantities in $K'$ :

\begin{aligned} u'_{\text{stone}}(t) &= u(t)-v_0 = gt-v_0 \\ u'_{\text{ground}} &= 0-v_0 = -v_0 \end{aligned} \tag{1}

Step 2The stone and the ground start at relative rest in

K'

The release instant in the two frames. In K the stone is at rest a height h above the ground. In K', falling at v_0, both the stone and the ground move upward at that one speed, so the two start at relative rest. — The release instant in the two frames. In $K$ the stone is at rest a height $h$ above the ground. In $K'$ , falling at $v_0$ , both the stone and the ground move upward at that one speed, so the two start at relative rest.

Evaluating the stone’s velocity (1) at the release instant $t=0$ ,

u'_{\text{stone}}(0) = g(0)-v_0 = -v_0,

which is exactly the ground’s velocity $u'_{\text{ground}}=-v_0$ found above. So at $t=0$ the relative velocity of the stone with respect to the ground in $K'$ is

u'_{\text{stone}}(0)-u'_{\text{ground}} = 0

(2)

even though neither body is individually at rest in $K'$ : both instantaneously share the single velocity $-v_0$ (a speed $v_0$ upward), which is exactly what the figure’s right-hand panel shows.

Step 3Integrate the relative motion in

K'

Differentiating $u'_{\text{stone}}(t)$ and $u'_{\text{ground}}$ from (1), the stone’s velocity relative to the ground in $K'$ is

v_{\text{rel}}(t) = u'_{\text{stone}}(t)-u'_{\text{ground}} = (gt-v_0)-(-v_0) = gt,

which vanishes at $t=0$ , consistent with (2). Integrating this relative velocity from release to the contact instant $t=T$ gives the relative displacement covered by then,

\int_0^T v_{\text{rel}}(t)\,dt = \int_0^T gt\,dt = \frac{1}{2}gT^2.

A Galilean transform translates both bodies’ positions by the same amount at any instant, so the separation between the stone and the ground at $t=0$ is still $h$ in $K'$ , exactly as in $K$ ; the relative displacement above must therefore equal $h$ ,

\frac{1}{2}gT^2 = h \;\Longrightarrow\; T=\sqrt{\frac{2h}{g}}.

Substituting back into $v_{\text{rel}}(t)=gt$ at $t=T$ gives the relative speed at contact.

v_0 = g\sqrt{\dfrac{2h}{g}} = \sqrt{2gh}

(3)

because the relative speed of approach between the stone and the ground is exactly the quantity the problem calls $v_0$ , and it cannot depend on which inertial frame it was computed in.

Check

A striking by-product confirms the bookkeeping: evaluating $u'_{\text{stone}}$ from (1) at the contact instant $t=T=v_0/g$ gives $u'_{\text{stone}}(T)=gT-v_0=v_0-v_0=0$ , so the stone arrives at the ground with zero velocity of its own in $K'$ . That is consistent, not contradictory: the ground itself is still moving upward at $v_0$ in $K'$ at that instant, so the relative velocity is $0-(-v_0)=v_0$ , matching (2)’s zero at the start and the boxed result at the end.

Remark

The formula $v_0=\sqrt{2gh}$ was never really a statement about the stone’s velocity in one particular frame; it is a statement about the relative velocity of approach between the stone and the Earth, and relative velocity is Galilean invariant, so the stone and the ground share a common velocity at the release instant in every inertial frame, $K$ included (both simply start at rest there). What is special about $K'$ is the other end of the motion: by the checkline above, $K'$ is the unique inertial frame in which the stone’s own velocity happens to vanish exactly at the moment of impact, which is why, viewed from inside $K'$ , the whole event reads as a fresh drop from rest, even though neither body is at rest in an absolute sense.