PHYS-1401: College Physics-I
CRN 55178
HOMEWORK PROBLEMS
Khalid Bukhari
HW-3
Chapter 3: VECTORS AND TWO DIMENSIONAL MOTION
PART-A: Hand in your answers in class on scantron on Wednesday 15 September-2010
1.
A catapult launches a large stone at a speed of 45.0 m/s at an angle of 55.0° with the horizontal.
What maximum height does the stone reach? (Neglect air friction.)
(a) 45.7 m
(b) 32.7 m
(c) 69.3 m
(d) 83.2 m
(e) 102 m
At maximum height (y =hmax), the vertical velocity of the stone will be zero. Thus,
v 2y  v 02 y  2ay (y) gives
hmax 
v 2y  v 02 y
2ay

  45 m s  sin 2  55.0 
0  v 02 sin 2 


 69.3 m
2  g 
2 9.80 m s2
2


and we see that choice (c) is the correct answer.
2.
A skier leaves the end of a horizontal ski jump at 22.0 m/s and falls 3.20 m before landing.
Neglecting friction, how far horizontally does the skier travel in the air before landing?
(a) 9.8 m
(b) 12.2 m
(c) 14.3 m
(d) 17.8 m
(e) 21.6 m
The skier has zero initial velocity in the vertical direction ( 0y = 0) and undergoes a vertical
displacement of y  3.20 m . The constant acceleration in the vertical direction is ay  g , so
we use y  v 0 y t 
1
2
a y t 2 to find the time of flight as
3.20 m  0 
1
9.80 m s2 t 2 or t 
2


2  3.20 m 
9.80 m s2
 0.808 s
During this time, the object moves with constant horizontal velocity v x  v 0 x  22.0 m s The horizontal distance
traveled during the flight is
 x  v x t   22.0 m s   0.808 s   17.8 m
which is choice (d).
3.
A cruise ship sails due north at 4.50 m/s while a coast guard patrol boat heads 45.0° north of
west at 5.20 m/s. What is the velocity of the cruise ship relative to the patrol boat?
(a) vx = 3.68 m/s; vy = 0.823 m/s
(b) vx = −3.68 m/s; vy = 8.18 m/s
(c) vx = 3.68 m/s; vy = 8.18 m/s
(d) vx = −3.68 m/s; vy = −0.823 m/s
(e) vx = 3.68 m/s; vy = 1.82 m/s
Choose coordinate system with north as the positive y-direction and east as the positive
x-direction. The velocity of the cruise ship relative to Earth is vCE  4.50 m s due north, with
components of (vCE )x  0 and (vCE) y  4.50 m s . The velocity of the patrol boat relative to
Earth is vPE  5.20 m s at 45.0° north of west, with components of
and
 vPE  x
  v PE cos 45.0     5.20 m s   0.707   3.68 m s
 vPE  y
  vPE sin 45.0  5.20 m s   0.707   3.68 m s
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PHYS-1401: College Physics-I
CRN 55178
Khalid Bukhari
HW-3
Thus, the velocity of the cruise ship relative to the patrol boat is vCP  vCE  vPE , which has components of
and
 vCP  x
  v CE  x   v PE  x  0   3.68 m s   3.68 m s
 vCP  y
  vCE  y   vPE  y  4.50 m s  3.68 m s   0.823 m s
Choice (a) is then the correct answer.
4.
An athlete runs three-fourths of the way around a circular track. Which of the following
statements is true?
(a) His average speed is greater than the magnitude of his average velocity.
(b) The magnitude of his average velocity is greater than his average speed.
(c) His average speed is equal to the magnitude of his average velocity.
(d) His average speed is the same as the magnitude of his average velocity if his instantaneous speed is
constant.
(e) None of statements (a) through (d) is true.
5.
A car moving around a circular track with constant speed
(a) has zero acceleration,
(b) has an acceleration component in the direction of its velocity,
(c) has an acceleration directed away from the center of its path,
(d) has an acceleration directed toward the center of its path, or
(e) has an acceleration with a direction that cannot be determined from the information given.
6.
A NASA astronaut hits a golf ball on the Moon. Which of the following quantities, if any,
remain constant as the ball travels through the lunar vacuum?
(a) speed
(b) acceleration
(c) velocity (d) horizontal component of velocity (e) vertical
component of velocity
7.
A baseball is thrown from the outfield toward the catcher. When the ball reaches its highest
point, which statement is true?
(a) Its velocity and its acceleration are both zero.
(b) Its velocity is not zero, but its acceleration is zero.
(c) Its velocity is perpendicular to its acceleration.
(d) Its acceleration depends on the angle at which the ball was thrown.
(e) None of statements (a) through (d) is true.
8.
A student throws a heavy red ball horizontally from a balcony of a tall building with an initial
speed v0. At the same time, a second student drops a lighter blue ball from the same balcony. Neglecting
air resistance, which statement is true?
(a) The blue ball reaches the ground first.
(b) The balls reach the ground at the same instant.
(c) The red ball reaches the ground first.
(d) Both balls hit the ground with the same speed.
(e) None of statements (a) through (d) is true.
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PHYS-1401: College Physics-I
CRN 55178
Khalid Bukhari
HW-3
9. John throws a baseball from the outfield from shoulder height, at an initial velocity of 29.4 m/s at an
initial angle of 30.0 with respect to the horizontal. The ball is in its trajectory for a total interval of 3.00
s before the third baseman catches it at an equal shoulder-height level. (Assume air resistance
negligible.) What is the ball’s horizontal displacement?
a. 76.4 m
b. 38.2 m
c. 57.3 m
d. zero
Vx = 29.4 Cos 30 = 25.46 m/s
Distance = Vx * t = 25.46 * 3 = 76.4 m
10.
A stone is thrown at an angle of 30 above the horizontal from the top edge of a cliff with an initial speed
of 12 m/s. A stop watch measures the stone’s trajectory time from top of cliff to bottom to be 5.6 s. How far out
from the cliff’s edge does the stone travel horizontally? (g = 9.8 m/s2 and air resistance is negligible)
a. 58 m
b. 154 m
c. 120 m
d. 197 m
Vx = 12 Cos 30 = 10.39 m/s
Distance = Vx *t = 10.39 * 5.6 = 58.2 m
11.
A boat moves at 10.0 m/s relative to the water. If the boat is in a river where the current is 2.00
m/s, how long does it take the boat to make a complete round trip of 1 000 m upstream followed by a 1
000-m trip downstream?
a. 200 s
b. 203 s
c. 208 s
d. 250 s
Time = Distance / velocity
T(downstream) = 1000m / (10+2) = 1000 / 12 = 83.3 s
T(upstream) = 1000m / (10-2) = 1000 / 8 = 125 s
Total time = 83.2 + 125 = 208.3 s
3
PHYS-1401: College Physics-I
CRN 55178
Khalid Bukhari
HW-3
PART-B: Hand in your solutions to the following questions in class, on Wednesday 15 September-2010.
Show the detailed calculations. Write your final answers in the box.
12. A ball is rolled horizontally off a table with an initial speed of 0.24 m/s. A stopwatch measures the ball’s
trajectory time from table to the floor to be 0.30 s. (g = 9.8 m/s2 and air resistance is negligible)
a) What is the height of the table?
b) How far away from the table does the ball land?
yf = yi + Vit + (1/2)at2
yf = 0 + 0 + (1/2) (-9.8) (0.3)2
yf = -0.441 m
Height of table = - 0.44 m
Distance = Vx * t
= 0.24 * 0.3 = 0.072 m
13. A fireman, 50.0 m away from a burning building, directs a stream of water from a fire hose at an angle of
30.0 above the horizontal. If the initial speed of the stream is 40.0 m/s, at what height will the stream of
water strike the building?
Time = distance / velocity
= 50 / (40 Cos 30) = 1.44 s
yf = yi + Vit + (1/2)at2
= 0 + (40 Cos 30)*(1.44) + (1/2) (-9.8)(1.44)2
= 18.6 m
14. Plane A is flying at 400 mph in the northeast direction relative to the earth. Plane B is flying at 500
mph in the north direction relative to the earth. What is the direction of motion of Plane B as observed
from Plane A?
VA = 400 Cos 45 i + 400 Sin 45 j
= 282.8 i + 282.8 j
VB = 500 j
Relative velocity = VB – VA = 282.8 i – 217.1 j
| V | = √( 282.8 2 – 217.1 2)
= 356 m/s
Direction = tan-1 (-217.1/282.8) = -37.5°
= 37.5 ° N of W
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