Problem Set 3.7:  1
 5 
7  9  11  13

15  19  22  21
 24  27  29  31
 32  Go up
 by To To Chong, Cooper
Spear, and Chris Murray Oct/2001
1. A car is driven 125 km west and then 65 km southwest. What is the displacement of the car form the point of origin (magnitude and direction)? Draw a diagram.
y1 = 0
x1 = 125km
y2 = 65cos45 = 46km
x2 = 65sin45 = 46km
x = x1 + x2 = 125km  46km = 171km
y = y1 + y2 = 46km
The displacement of the car is: D^{2} = x^{2} + y^{2} è D = ((46 km)^{2} + (171 km)^{2})^{1/2 }= 177km
tan@ = 46/171 = 0.27 ==> @ =15degree South of West
5. Graphically determine the resultant of the following three vector displacements: (1) 24 m, 30° north of east; (2) 28m 37° east of north; and (3) 20 m, 50° west of south?
Here is what those vectors look like:
Now, they are asking us to add these vectors graphically, which would look like:
Where the black vector is the resultant.
But we can do better than that
Vector 1 has these components, both positive:
x1 = 24cos30 = 20.78m
y1 = 24sin30 = 12m
Vector 1 = 20.78 m x + 12 m y
Vector 2 has these components, both positive:
x2 = 28sin37 = 16.85m
y2 = 28cos37 = 22.36m
Vector 2 = 16.85 m x + 22.36 m yVector 3 has these components, both negative:
x3 = 20sin50 = 15.32m
y3 = 20cos50 = 12.86m
Vector 3 = 15.32 m x + 12.86 m yThe sum of these three would be:
Vector X direction Y direction 1 20.78 m x + 12 m y 2 16.85 m x + 22.36 m y 3 15.32 m x + 12.86 m y Sum: 22.31 m x + 21.5 m
The resultant is: D = (x^{2} + y^{2})^{1/2 } = 31m
The angle is tan1(21.5/22.31) = 44 degrees North of East
(Table of contents)
7. Figure 333 shows two vectors, A and B, whose magnitudes are A=8.31 units and B=5.55 units. Determine C if (a) C=A+B, (b) C=AB, (c) C=BA. Give the magnitude and direction for each.
This is baby hard. A, since it is to the right would be 8.31 x + 0 y, and B, since it is to the left, would be 5.55 x + 0 y.
So A+B:
A: 8.31 x + 0 y + B: 5.55 x + 0 y = 2.76 x + 0 y AB:
A: 8.31 x + 0 y  B: 5.55 x + 0 y = 13.86 x + 0 y
BA:
B: 5.55 x + 0 y  A: 8.31 x + 0 y = 13.86 x + 0 y
9. A airplane is traveling 785 km/h in a direction 38.5° west of north (Fig. 334). (a) Find the components of the velocity vector in the northerly and westerly directions. (b) How far north and how far west has the plane traveled after 3.00 h?
Let's break the vector into components first:
The Westward component is found with the Sin, and the Northerly component is found with Cos:
V_{West} = (785 km/hr)sin(38.5^{o}) = 488.67 km/hr = 489 km/hr
V_{North} = (785 km/hr)cos(38.5^{o}) = 614.35 km/hr = 614 km/r
To find out how far in 3.00 hr, just multiply by time:
v = s/t, s = vt:
S_{West} = (488.67 km/hr)(3.00 hr) = 1470 km
V_{North} = (614.35 km/hr)(3.00 hr) = 1840 km
(Table of contents)
11.Three vectors are shown in Fig. 335. Their magnitudes are given in arbitrary units. Determine the sum of the three vectors. Give the resultant in terms of (a) components, (b) magnitudes and angle with x axis.
First let's find the components:
Vector A has two positive components, the x is found with Cos since it is adjacent to the angle, and the y is found with Sin:
A = +(66.0)Cos(28.0^{o}) + (66.0)Sin(28.0^{o})
A = 58.27 x + 30.99 y
Vector B has a negative x component (goes to the left), and a positive y component (goes up). The x is adjacent to the angle, and so we use Cos, and the y is again opposite, so we use Sin:B = (40.0)Cos(56.0^{o}) + (40.0)Sin(56.0^{o})
B = 22.37 x + 33.16 y
Vector C is easy, it is all in the y direction (it goes down), so
C = 0 x + 46.8 y
So let's add them up like we're told:
Vector X direction Y direction A 58.27 x + 30.99 y B 22.37 x + 33.16 y C 0 x + 46.8 y Sum: 35.9 x + 17.35 y
Now we are ready to turn it into an Angle Magnitude vector:
Draw the new vector:
Find the magnitude:
hyp = Ö{(35.9)^{2} + (17.35)^{2}} = 39.9
The angle between the x axis (blue) and the vector (black) is the inverse tangent of opposite over adjacent:
q =Tan^{1}{17.35/35.9} = 25.8^{o}
(Table of contents)
13. (a) Given the vectors A and B shown in Fig. 335, determine BA. (b) Determine AB without using your answer in (a). Then compare your results and see if they are opposite.
First let's find the components:
Vector A has two positive components, the x is found with Cos since it is adjacent to the angle, and the y is found with Sin:
A = +(66.0)Cos(28.0^{o}) + (66.0)Sin(28.0^{o})
A = 58.27 x + 30.99 y
Vector B has a negative x component (goes to the left), and a positive y component (goes up). The x is adjacent to the angle, and so we use Cos, and the y is again opposite, so we use Sin:B = (40.0)Cos(56.0^{o}) + (40.0)Sin(56.0^{o})
B = 22.37 x + 33.16 y
So first find B  A:
B  A is the same as B + (A), and since A = 58.27 x + 30.99 y, A = 58.27 x + 30.99 y
So let's add A to B:
Vector X direction Y direction A 58.27 x + 30.99 y B 22.37 x + 33.16 y Sum: 80.64 x + 2.17 y Now we are ready to turn it into an Angle Magnitude vector:
Draw the new vector:
Find the magnitude:
hyp = Ö{(80.64)^{2} + (2.17)^{2}} = 80.7
The angle between the x axis (blue) and the vector (black) is the inverse tangent of opposite over adjacent:
q =Tan^{1}{2.17/80.64} = 1.54^{o}
Now let's find A  B:
A  B is the same as A + (B), and since B = 22.37 x + 33.16 y, B = +22.37 x + 33.16 y
So let's add A to B:
Vector X direction Y direction A 58.27 x + 30.99 y B +22.37 x + 33.16 y Sum: 80.64 x + 2.17 y Now we are ready to turn it into an Angle Magnitude vector:
Draw the new vector:
Find the magnitude:
hyp = Ö{(80.64)^{2} + (2.17)^{2}} = 80.7
The angle between the x axis (blue) and the vector (black) is the inverse tangent of opposite over adjacent:
q =Tan^{1}{2.17/80.64} = 1.54^{o}
And, by golly, this is exactly opposite B  A.
(Table of contents)
15.For the vectors shown in Fig, 335, determine (a) CAB, (b) 2A3B+2C.
First let's find the components:
Vector A has two positive components, the x is found with Cos since it is adjacent to the angle, and the y is found with Sin:
A = +(66.0)Cos(28.0^{o}) + (66.0)Sin(28.0^{o})
A = 58.27 x + 30.99 y
Vector B has a negative x component (goes to the left), and a positive y component (goes up). The x is adjacent to the angle, and so we use Cos, and the y is again opposite, so we use Sin:B = (40.0)Cos(56.0^{o}) + (40.0)Sin(56.0^{o})
B = 22.37 x + 33.16 y
Vector C is easy, it is all in the y direction (it goes down), so
C = 0 x + 46.8 y
So let's put them into a happy fun table:
Vector X direction Y direction A 58.27 x + 30.99 y B 22.37 x + 33.16 y C 0 x + 46.8 y Our first task is to find C  A  B, (Which spells CAB.....Coincidence??? I THINK NOT!!!!!!!!)
I am just going to add C, A and B together:
Vector X direction Y direction A 58.27 x + 30.99 y B +22.37 x + 33.16 y C 0 x + 46.8 y Sum: 35.9 x + 110.95 y
Now we are ready to turn it into an Angle Magnitude vector:
Draw the new vector:
Find the magnitude:
hyp = Ö{(35.9)^{2} + (110.95)^{2}} = 117
The angle between the x axis (blue) and the vector (black) is the inverse tangent of opposite over adjacent:
q =Tan^{1}{110.95/35.9} = 72.1^{o}
Our second task is to find 2A  3B + 2C,
I am just going to add 2A, 3B, and 2C together:
Since A = 58.27 x + 30.99 y,
2A = 2(58.27 x + 30.99 y) = 116.54 x + 61.98 y
B = 22.37 x + 33.16 y,
3B = 3(22.37 x + 33.16 y) = 67.11 x + 99.48 y
C = 0 x + 46.8 y,
2C = 2(0 x + 46.8 y) = 0 x + 93.6 y
So let's add them together:
Vector X direction Y direction 2A 116.54 x + 61.98 y 3B 67.11 x + 99.48 y 2C 0 x + 93.6 y Sum: 183.65 x + 131.1 y
Now we are ready to turn it into an Angle Magnitude vector:
Draw the new vector:
Find the magnitude:
hyp = Ö{(183.65)^{2} + (131.1)^{2}} = 226
The angle between the x axis (blue) and the vector (black) is the inverse tangent of opposite over adjacent:
q =Tan^{1}{131.1/183.65} =35.5^{o}
(Table of contents)
19. A tiger leaps horizontally form a 7.5mhigh rock with a speed of 4.5 m/s. How far from the base of the rock will she land?
Horizontal Vertical s =
u = 4.5 m/s
v = "
a = 0
t =s = 7.5 m
u = 0 (Purely horizontal)
v = ? (don't care!)
a = 9.8 m/s/s
t = ?Use s = ut + ^{1}/_{2}at^{2 }7.5 = ^{1}/_{2}(9.8)t^{2 }
t = 1.2372 sUse: v_{av = }Ds/Dt, as we now know t
4.5 m/s = s/(1.2372s)
s = 5.567 m = 5.6m
22. Romeo is chucking pebbles gently up to Juliet's window, and he wants the pebbles to hit the window with only a horizontal component of velocity. He is standing at the edge of a rose garden 8.0 m below her window and 9.0 m from the base of the wall (Fig. 338). How fast are the pebbles going when they hit her window?
Horizontal Vertical s = 9.0 m
u = ???
v = "
a = 0
t =s = 8.0 m
u = ?
v = 0 (final is purely horizontal)
a = 9.8 m/s/s
t = ?From the vertical we've got: t = 1.278s Use v^{2} = u^{2 + }2as^{ }0 = u^{2 + }2(9.8)(8.0)
u = 12.522m/sThe horizontal velocity is: x = ut = vt v = x/t = 9/1.278 = 7m/s
v = u + at t = u/g = 12.522/9.8 = 1.278s
As the vertical final velocity is zero, the velocity of the pebble when it hits the windows is: V = (V_{x}^{2} + V_{y}^{2})^{1/2} = (0^{2} + 7^{2})^{1/2} = 7m/s
21. A fire hose held near the ground shoots water at a speed of 6.5 m/s. At what angle(s) should the nozzle point in order that the water land 2.0 m away (Fig. 337) ? Why are there different angles?
use the range equation (range = v^{2}sin(2q)/g)
The necessary angle is:
sin2@ = (Rg)/v^{2} = (2 * 9.8)/6.5^{2} = 0.464 ==> 2@ = 27.64^{0} è @ = 14^{0}
Or @ = 90^{0} – 14^{0} = 76^{0}
We've got two different angle because:
 At the larger angle the water has a smaller horizontal velocity but spends more time in the air, because of the larger initial vertical velocity, thus the horizontal displacement is the same for the two angles.
24. A ball is thrown horizontally from the roof of a building 56 m tall lands 45 m from the base . What was the ball's initial speed?
Horizontal Vertical s = 45 m
u = ???
v = "
a = 0
t = ?s = 56 m
u = 0 (purely horizontal)
v = ?
a = 9.8 m/s/s
t = ?The ball's initial speed is horizontal. So we have: s = vt = ut
==> u = s/t = 45/3.38 = 13.31m/s
Use s = ut + ^{1}/_{2}at^{2 t = (2s/g)1/2 = (2 * 56/9.8)1/2 = 3.38s }
27. A ball thrown horizontally at 22.2 m/s from the roof of a building lands 36.0 m from the base of the building. How high is the building?
Choose the coordinate system Oxy like this, then we have:
Horizontal Vertical s = 36.0 m
u = 22.2 m/s
v = "
a = 0
t = ?s = ????
u = 0 (purely horizontal)
v = ?
a = g = 9.8 m/s/s
t = ?x = vt = ut t = x/u = 36/22.2 = 1.6216s
The height of the building is: s = ut + 1/2at^{2}
y_{0} = 0
s = 1/2gt^{2} = ½ * 9.8 * 1.6216^{2} = 12.9m
29. Determine how much farther a person can jump on the Moon as compared to the Earth if the takeoff speed and angle are the same. The acceleration due to gravity on the Moon is onesixth what is on the earth.
Use the range equation (range = v^{2}sin(2q)/g)
Suppose the takeoff speed is u, the angle is @.
Due to the exercise, we have this: g_{M} = 1/6g_{E}
R_{E} = (u^{2}sin2@)/g_{E}
R_{M} = (u^{2}sin2@)/g_{M}
R_{M}/R_{E} = g_{E}/g_{M} = 6
So it is 6 times farther in the moon.
31. The pilot of an airplane traveling160 km/h wants to drop supplies to flood victims isolated on a patch of land 160 m below. The supplies should be dropped how many how many seconds before the plane is directly overhead?
Horizontal Vertical s = 36.0 m
u = 160/3.6 = 44.4444 m/s (not used)
v = "
a = 0
t = ?s = 160 m
u = 0 (purely horizontal)
v = ?
a = 9.8 m/s/s
t = ??????Use s = ut + ^{1}/_{2}at^{2 160 = 0 + 1}/_{2}(9.8)t^{2}
So t = 5.714 s = 5.7 s
32. A hunter aims directly at a target (on the same level) 120 m away. (a) If the bullet leaves the gun at a speed of 250 m/s, by how much will it miss the target? (b) At what angle should the gun be aimed so the target will be hit?
a.
Horizontal Vertical s = 120 m
u = 250 m/s
v = "
a = 0
t = ?s = ????
u = 0 (purely horizontal)
v = ?
a = 9.8 m/s/s
t = ?Use:
v_{av = }Ds/Dt250 m/s = 120 m/t,
t = .48 sUse s = ut + ^{1}/_{2}at^{2}
s = 0 + ^{1}/_{2}(9.8)(.48)^{2 } = 1.129 = 1.1 m
b. use the range equation (range = v^{2}sin(2q)/g)
so q = sin^{1}(g(range)/v^{2})/2 = .539 = .54^{o} above horiz