Physics quiz question: airplane trading altitude for speed

Physics 205A Quiz 4, fall semester 2013
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Comprehensive Problem 6.79

"crop duster"
cdn-pix
flic.kr/p/8ciFEK

A crop duster airplane[*] is flying at its cruise speed of 64 m/s (143 mph). It descends 20 m in order to "trade altitude for speed."[**] Ignore friction/drag, and work done by the engine during this process. The final speed of the airplane after its descent is:
(A) 20 m/s.
(B) 67 m/s.
(C) 84 m/s.
(D) 110 m/s.

[*] wiki.pe/Air_Tractor_AT-400.
[**] Planes, Walt Disney Pictures (2013).

Correct answer (highlight to unhide): (B)

The energy transfer-balance equation is given by:

W_nc = ∆KE_tr + ∆PE_grav + ∆PE_elas,

where W_nc = 0 (no external gains/losses of mechanical energy due to the engine or friction/drag), and ∆PE_elas = 0 (no springs involved), such that the remaining terms in the equation are:

0 = ∆KE_tr + ∆PE_grav,

0 = (1/2)·m·(v_f² – v₀²) + m·g·(y_f – y₀).

The mass m cancels out, and solving for the final speed v_f:

– g·(y_f – y₀) = (1/2)·(v_f² – v₀²),

v₀² – 2·g·(y_f – y₀) = v_f²,

√(v₀² – 2·g·(y_f – y₀)) = v_f,

√((64 m/s)² – 2·(9.80 m/s²)·((0 m) – (20 m))) = v_f,

then v_f = 66.9925369 m/s, or to two significant figures, 67 m/s.

(Response (A) is √(–2·g·(y_f – y₀)); response (C) is v₀ + √(–2·g·(y_f – y₀)); response (D) is √(–v₀·g·(y_f – y₀)).)

Sections 70854, 70855, 73320
Exam code: quiz04iSs5
(A) : 0 students
(B) : 23 students
(C) : 41 students
(D) : 2 students

Success level: 35%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.66