Cuesta College, San Luis Obispo, CA
Scott Gaunson for the "How Ridiculous" YouTube channel threw a golf ball up towards the top platform of the Gravity Discovery Centre and Observatory's Leaning Tower of Gingin, near Perth, Australia[*']. Video analysis shows that from the moment it was released from his hand (with an initial upwards speed), the golf ball took 1.8 s to travel 45 m upwards to the top platform, where it still had an upwards speed of 15 m/s. Neglect air resistance. Choose up to be the +y direction. At the moment it left his hand, the golf ball had an initial speed of:
(A) 25 m/s.
(B) 27 m/s.
(C) 33 m/s.
(D) 43 m/s.
Correct answer (highlight to unhide): (C)
The following quantities are given (or assumed to be known):
(t0 = 0 s),
(y0 = 0 m),
y = +45 m (level of the top platform above the throw release),
vy = +15 m/s (moving upwards at the level of top platform),
t = 1.8 s (time when at the level of top platform),
ay = –9.80 m/s2.
So in the equations for constant acceleration motion in the vertical direction, there are no quantities that are unknown, and only one to be explicitly solved for:
vy = v0y + ay·t,
y = (1/2)·(vy + v0y)·t,
y = v0y·t + (1/2)·ay·(t)2,
vy2 = v0y2 + 2·ay·y.
Note that as the quantity v0y to be solved for appears as the only unknown in every one of the equations above, with all other quantities given (or assumed to be known), then any one of these equations could be used to solve for the initial velocity. Here we only demonstrate how the first equation is used:
vy = v0y + ay·t,
+15 m/s = v0y + (–9.80 m/s2)·(1.8 s),
v0y = +15 m/s –(–9.80 m/s2)·(1.8 s) = +15 m/s + 17.64 m/s = +32.64 m/s,
or to two significant figures, the initial speed of the golf ball was 33 m/s (in the upwards direction).
(Response (A) is y/t; response (B) is vf·t; response (D) is vf·t – (1/2)·ay·(t2).)
Sections 70854, 70855
Exam code: quiz02HwRd
(A) : 5 students
(B) : 5 students
(C) : 41 students
(D) : 1 student
Success level: 79%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.31