Physics quiz question: S.M.T.H. record toss

Physics 205A Quiz 2, fall semester 2013
Cuesta College, San Luis Obispo, CA

"S.M.T.H. (Send Me To Heaven)" (excerpt)
petrsvar
youtu.be/pXr84w1cWIU

S.M.T.H. (Send Me To Heaven) is an Android application that registers how high a device is thrown up in the air.[*] The world record holder[**] was a device thrown up to a maximum height of 44.19 m from where it was released. Neglect air resistance. Choose up to be the +y direction. This device was released with an initial upwards velocity of:
(A) +1.36 m/s.
(B) +3.00 m/s.
(C) +9.02 m/s.
(D) +29.4 m/s.

[*] play.google.com/store/apps/details?id=com.carrotpop.www.smth.
[**] "World Top 10: 1. LIATO, 44.19 m," as of August 8, 2013.

Correct answer (highlight to unhide): (D)

The following quantities are given (or assumed to be known):

(t₀ = 0 s),
(y₀ = 0 m),
y = +44.19 m (at maximum height above where it was released),
v_y = 0 m/s (at maximum height above where it was released),
a_y = –9.80 m/s².

So in the equations for constant acceleration motion in the vertical direction, the following quantities are unknown, or are to be explicitly solved for:

v_y = v_0y + a_y·t,

y = (1/2)·(v_y + v_0y)·t,

y = v_0y·t + (1/2)·a_y·(t)²,

v_y² = v_0y² + 2·a_y·y.

With the unknown quantity v_0y to be solved for appearing in the fourth equation, with all other quantities given (or assumed to be known), then:

v_y² = v_0y² + 2·a_y·y,

v_0y² = v_y² – 2·a_y·y = (0 m/s)² – 2·(–9.80 m/s²)·(+44.19 m) = 866.124 m²/s²,

v_0y = ±29.42998471 m/s,

or to three significant figures, the initial upwards velocity (choosing the positive root) is +29.4 m/s.

(Response (A) is 2·sqrt(y)/a_y; response (B) is t = sqrt(2·y/a_y); response (C) is 2·y/a_y.)

Sections 70854, 70855, 73320
Exam code: quiz02p3nG
(A) : 2 students
(B) : 5 students
(C) : 9 students
(D) : 53 students

Success level: 77%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.46