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Material Type: Quiz; Professor: Antoniewicz; Class: ENGINEERING PHYSICS I; Subject: Physics; University: University of Texas - Austin; Term: Spring 2011;
Typology: Quizzes
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Version 036/AACBA – QTest #3 – Antoniewicz – (57420) 1
This print-out should have 4 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering.
001 (part 1 of 4) 10.0 points A plane is flying horizontally with speed 287 m/s at a height 1870 m above the ground, when a package is dropped from the plane. The acceleration of gravity is 9.8 m/s^2. Neglecting air resistance, when the package hits the ground, the plane will be
Explanation: Both have the same velocity at the time of release. Gravitational acceleration does not change horizontal velocity, so the plane will be directly above the package.
002 (part 2 of 4) 10.0 points What is the horizontal distance from the re- lease point to the impact point?
Correct answer: 5606.67 m.
Explanation:
For the vertical fall, h =
g t^2 , or t = √ 2 h g
. The horizontal distance traveled is
x = v t
= v
2 h g
= (287 m/s)
2 (1870 m) 9 .8 m/s^2 = 5606.67 m.
003 (part 3 of 4) 10.0 points A second package is thrown downward from the plane with a vertical speed v 1 = 72 m/s. What is the magnitude of the total velocity of the package at the moment it is thrown as seen by an observer on the ground?
Correct answer: 295.894 m/s.
Explanation: The velocity is the vector sum of the vertical and horizontal components of velocity as seen from the ground. Hence the scalar speed is
s =
v^2 + v^21
=
(287 m/s)^2 + 72 m/s^2 = 295.894 m/s.
004 (part 4 of 4) 10.0 points What horizontal distance is traveled by this package?
Version 036/AACBA – QTest #3 – Antoniewicz – (57420) 2
Correct answer: 3881.49 m.
Explanation: The time of the vertical fall is now deter- mined by
h = v 1 t +
g t^2
0 =
g t^2 + v 1 t − h
t =
−v 1 +
v^21 + 4 ( 12 ) g h 2 12 g
−v 1 +
v^21 + 2 g h g = 13.5243 s.
The horizontal distance is
x = v t = (287 m/s) (13.5243 s) = 3881.49 m.