Physics 205A Quiz 3, spring semester 2009
Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 3.46
A ball is thrown off the edge of a cliff with an initial horizontal velocity. It lands on the ground below 1.9 s after it was thrown, at a horizontal distance of 32 m from the base of the cliff. Neglect air resistance. Choose up to be the +y direction. The initial (horizontal) speed of the ball is:
(A) 5.2 m/s.
(B) 17 m/s.
(C) 19 m/s.
(D) 61 m/s.
Correct answer (highlight to unhide): (B)
The following quantities are given (or assumed to be known):
(x0 = 0 m),
(y0 = 0 m),
(t0 = 0 s),
t = 1.9 s,
x = +32 m,
v0y = 0 m/s,
ay = –9.80 m/s2.
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 v0x to be solved for appearing in the fourth equation, with all other quantities given (or assumed to be known), then the initial horizontal velocity can be solved for:
v0x = x/t = +16.84 m/s,
of which the magnitude (speed) is 17 m/s to two significant digits.
(Response (A) is the result of (9.80 m/s2)/t; response (C) is (9.80 m/s2)·t, which would be the magnitude of the vertical (downwards) velocity of the ball 1.9 s after it was thrown; while response (D) is x·t.)
Student responses
Sections 30880, 30881
(A) : 1 student
(B) : 36 students
(C) : 11 students
(D) : 0 students
Success level: 75%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.43
Subscribe to:
Post Comments (Atom)
No comments:
Post a Comment