(a) Free-fall time $t=\sqrt{\dfrac{2h}{g+w}}=0.7$ s. (b) In the shaft frame the bolt ends up $0.7$ m below its start, but the path it travels is $s=\lvert\Delta y\rvert+2\dfrac{v_0^{2}}{2g}=1.3$ m.
(ii) Problem Statement
"An elevator car whose floor-to-ceiling distance is $h=2.7$ m starts ascending with constant acceleration $w=1.2$ m/s$^2$. A time $t_0=2.0$ s after the start, a bolt begins falling from the ceiling of the car. Find (a) the bolt's free-fall time; (b) the displacement and the distance covered by the bolt during the free fall, in the reference frame fixed to the elevator shaft."
The key idea: inside the rising lift the bolt feels gravity plus an extra downward pull (a pseudo-force), so it drops as if gravity were $g+w$. That makes the falling time short. But in the shaft, the bolt is first flung upward at the lift's speed, so its path is longer than its net drop.
(iii) The Big Idea in Plain Words
A lift speeds upward. A bolt is stuck to its ceiling. After a while the bolt comes loose and falls to the lift floor.
Symbols: $h$ is the ceiling height above the floor. $w$ is the lift's upward acceleration. $t_0$ is the wait before the bolt drops. $g$ is gravity's pull, $9.8$ m/s$^2$. $v_0$ is the lift's speed when the bolt lets go.
Two different questions hide here. "How long to fall?" is easiest inside the lift, where the bolt just drops $h$ under a stronger pull $g+w$. "How far does the bolt actually travel?" is a shaft question, where the bolt first flies up, stops, then comes down — so the path beats the net drop.
(iv) The Solution Viewed in Each Frame
1. Elevator Frame (rides with the car) — best for part (a)
Why: the car speeds up, so inside it a pseudo-force $mw$ points down, adding to gravity.
The bolt's downward acceleration here is $g+w$. It starts at rest relative to the car (it was stuck to the ceiling).
Why: it must fall the whole height $h$ from rest.
$h=\tfrac12(g+w)t^{2}\Rightarrow t=\sqrt{\dfrac{2h}{g+w}}=\sqrt{\dfrac{2(2.7)}{9.8+1.2}}=0.7$ s.
2. Ground / Shaft Frame (stationary observer) — best for part (b)
Why: when the bolt lets go it already shares the lift's upward speed.
$v_0=wt_0=1.2\times2.0=2.4$ m/s, upward. After release only gravity acts, so acceleration is $g$ down.
Why: net move in the fall time is up minus down.
$\Delta y=v_0t-\tfrac12 gt^{2}=2.4(0.7)-\tfrac12(9.8)(0.7)^{2}=-0.7$ m. The minus means it ends $0.7$ m below its start point.
Why: the bolt first rises, so the path is longer than the net move.
Rise height $=\dfrac{v_0^{2}}{2g}=\dfrac{2.4^2}{19.6}=0.29$ m. Total path $s=\lvert\Delta y\rvert+2\dfrac{v_0^{2}}{2g}=0.7+0.59=1.3$ m. (The free-fall time is still $0.7$ s — the floor rises to meet the bolt, closing the gap $h$ at relative rate set by $g+w$.)
3. Centre-of-Mass Frame
Does not apply. The lift is driven by an outside motor (huge effective mass); this is not a two-body momentum problem.
4. Medium Frame
Does not apply. There is no air drag or fluid in the problem.
- Changes: the bolt's path — a straight drop of $h$ inside the lift, a little up-then-down arc in the shaft.
- Same: the free-fall time $0.7$ s (both frames agree on when bolt meets floor). The displacement $0.7$ m and path $1.3$ m are shaft-frame answers by definition.
(v) The Solution by Each Method
Method 1: Relative-Motion Method (fastest for the time)
- Why: use the bolt's motion relative to the floor.
Relative acceleration $=g+w$ (gravity plus the floor rising at $w$). Relative start speed $=0$. - Why: close the gap $h$.
$t=\sqrt{\dfrac{2h}{g+w}}=0.7$ s.
Method 2: Force Method (FBD)
- Why: after release the bolt has only its weight on it.
Shaft frame: $ma=-mg\Rightarrow a=-g$ (projectile). Lift frame: add pseudo-force $-mw$, so $a_{\text{rel}}=-(g+w)$. - Why: integrate the constant acceleration.
Lift frame gives $t=\sqrt{2h/(g+w)}=0.7$ s; shaft frame gives $\Delta y=-0.7$ m and path $1.3$ m.
Method 3: Graph Method (v–t and y–t)
- Why: on a shaft $y$–$t$ plot the bolt is a parabola and the floor is a parabola; they cross when the bolt lands.
They meet $0.7$ s after release. The bottom panel draws both curves and the crossing. - Why: on a $v$–$t$ plot, area is distance; the bolt's speed crosses zero at the top.
The two triangle areas (up then down) add to the path $1.3$ m, while they partly cancel for the displacement $0.7$ m.
Method 4: Calculus Method
- Why: write positions and solve.
Bolt (shaft): $y_b=y_R+v_0\tau-\tfrac12 g\tau^2$. Floor: $y_f=\tfrac12 w(t_0+\tau)^2$. Set $y_b=y_f$. - Why: the gap equation reduces to $h=\tfrac12(g+w)\tau^2$.
So $\tau=0.7$ s, and the path is $\int_0^{\tau}\lvert v_b\rvert\,d\tau=1.3$ m.
Method 5: Energy Method
Does not suit the time question (energy gives speeds, not durations). It confirms the landing speed, but the clean route is kinematics.
Fastest: Method 1 (lift frame). Most reliable in an exam: Method 4 (positions never lie). All give $t=0.7$ s, $\Delta y=0.7$ m down, path $1.3$ m.
(vi) Parameter Table
| Symbol | Meaning | Unit | Live value |
|---|
(vii) Common Exam Traps
- Using plain $g$ for the fall time. Inside the rising lift the pull is $g+w$, so the time is shorter: $\sqrt{2h/(g+w)}$, not $\sqrt{2h/g}$.
- Forgetting the bolt starts moving up. At release it shares the lift's speed $v_0=wt_0$, so in the shaft it first rises.
- Mixing up displacement and distance. The net move is $0.7$ m down; the path travelled is $1.3$ m. They are different because the bolt reverses direction.
- Thinking the floor stays put. In the shaft the floor also accelerates up, which is why the gap still closes in $0.7$ s.
- Adding $w$ to the fall in the shaft frame. In the shaft the bolt feels only $g$; the $w$ belongs to the floor's motion (or to the pseudo-force inside the lift).
Verified: the live simulation, both frames and all methods agree — $t=\sqrt{2h/(g+w)}=0.7$ s, $\Delta y=-0.7$ m, $s=1.3$ m, with $h$ in m, $w,g$ in m/s$^2$, $t_0,t$ in s.