Physics 205A Quiz 3, fall semester 2008
Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Comprehensive Problem 3.74(c)
A ball is kicked off the edge of a cliff with an initial speed of 18.0 m/s, 10.0° above the horizontal. Neglect air resistance. Choose up to be the +y direction. How long would it take for the ball to fall to the ground 22.0 m below?
(A) 1.82 s.
(B) 2.12 s.
(C) 2.24 s.
(D) 2.46 s.
Correct answer (highlight to unhide): (D)
The following quantities are given (or assumed to be known):
(x0 = 0),
(y0 = 0),
(t0 = 0 s),
y = –22.0 m,
ay = –9.80 m/s2,
where the initial horizontal and vertical velocity components of the initial velocity vector are:
v0x = (18.0 m/s)·cos(+10°) = +17.7 m/s,
v0y = (18.0 m/s)·sin(+10°) = +3.13 m/s.
So in the equations for projectile motion, the following quantities are unknown, or are to be explicitly solved for:
x = v0x·t,
vy - v0y = ay·t,
y = (1/2)·(vy + v0y)·t,
y = v0y·t + (1/2)·ay·(t)2,
vy2 - v0y2 = 2·ay·y.
With the unknown quantity t to be solved for appearing in the fourth equation, with all other quantities given (or assumed to be known), then it becomes a quadratic equation:
y = v0y·t + (1/2)·ay·(t)2,
0 = –y + v0y·t + (1/2)·ay·(t)2,
where the quadratic formula terms are "c" = –(–22.0 m) = +22.0 m; "b" = +3.13 m/s, and "a" = "–4.90 m/s2, resulting in the roots:
t = –1.82 s, +2.46 s,
of which the positive root (response (D)) is the sole realistic answer, given the initial conditions.
(Response (B) is the time t = sqrt(2·y/ay) it would take for the ball to fall to the ground if it were released from rest; response (A) is the time it would take for the ball to fall to the ground if it were thrown at an angle of 10.0° below the horizontal; while response (C) is merely –(y)/ay).
Student responses
Sections 70854, 70855
(A) : 7 students
(B) : 11 students
(C) : 13 students
(D) : 14 students
Success level: 37%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.73
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