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NAME ______________________________________ DATE _______________ CLASS ____________________

Two-Dimensional Motion and Vectors

Problem E
PROJECTILES LAUNCHED AT AN ANGLE
PROBLEM

The narrowest strait on earth is Seil Sound in Scotland, which lies between the mainland and the island of Seil
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0 m
wide
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What is the minimum initial speed
that would allow the athlete to clear the gap? Neglect air resistance
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PLAN

Given:

∆x = 6
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81 m/s2

Unknown:

1
...
00 m

Copyright © by Holt, Rinehart and Winston
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Choose the equation(s) or situation: The horizontal component of the athlete’s
velocity, vx , is equal to the initial speed multiplied by the cosine of the angle, q,
which is equal to the magnitude of the horizontal displacement, ∆x, divided by the
time interval required for the complete jump
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The time
required for this to occur is half the time necessary for the total jump
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΂

΃

∆x
−ay
vi sin q = ᎏᎏ 
2 v cos q
i
−ay ∆x
2 sin q cos q

vi 2 = ᎏᎏ
vi =

Ί๶๶

−ay∆x
ᎏᎏ
2 sin q cos q

Problem E

25

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NAME ______________________________________ DATE _______________ CLASS ____________________
3
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vi =

4
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81 m/s2)(6
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9 m/s

By substituting the value for vi into the original equations, you can determine the
time for the jump to be completed, which is 0
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From this, the height of the
jump is found to equal 1
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ADDITIONAL PRACTICE
1
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24 m
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) Suppose Brian threw the spear at a 35
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What was the initial speed of the spear?
2
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In 1981 in Utah, she sent an arrow a horizontal distance of
9
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What was the speed of the arrow at the top of the flight if the
arrow was launched at an angle of 45
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In 1989 during overtime in a high school basketball game in Erie, Pennsylvania, Chris Eddy threw a basketball a distance of 27
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If the shot was made at a 50
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What was the initial speed of the brick?
b
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If Capes threw the brick straight up with the speed found in (a),
what would be the maximum height the brick could achieve?
5
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5 m
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0°
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Michael Hout of Ohio can run 110
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9 s at an average speed of 5
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What makes this interesting is that he juggles three
balls as he runs the distance
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At what
angle, q , must the ball be thrown? (Hint: Consider horizontal displacements for Hout and the ball
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Allrights reserved
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In 1978, Geoff Capes of the United Kingdom won a competition for
throwing 5 lb bricks; he threw one brick a distance of 44
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Suppose
the brick left Capes’ hand at an angle of 45
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NAME ______________________________________ DATE _______________ CLASS ____________________

7
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42 m/s at an angle of 55
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If the jump
lasted 1
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Allrights reserved
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Measurements made in 1910 indicate that the common flea is an impressive jumper, given its size
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2 m/s, and
that it leaps at an angle of 21° with respect to the horizontal
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16 s, what is the magnitude of the flea’s horizontal displacement?
How high does the flea jump?

Problem E

27

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Givens

Solutions
vy 2 = 2ay ∆y

8
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95 m
vx = 3
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81 m/s2

v = vx 2 +ෆෆ2 = ෆay ∆y
vx2 + 2ෆ
ෆෆෆ v y ෆ
v = ෆෆ2)(−9
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95ෆ
(3
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0 m2/s2ෆෆ8
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3ෆ2/s2 = 6
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81 m/s2)(−1
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0 m/s

΂΃

΂

΃

΂

΃

q = 64° below the horizontal

Additional Practice E
1
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24 m
q = 35
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81 m/s2

∆x = vi (cos q)∆t
∆x
∆t = 
vi(cos q)

II

∆x
1
vi (sin q) = − ay 
2 vi(cosq)

΄ ΅
m/s )(201
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8135
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0°)
(2)(sin
2(sin
2

y

i

vi = 45
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∆x = 9
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0°

vi =

Copyright © by Holt, Rinehart and Winston
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ay = −g = −9
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81 m/s2)(9
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0°)(cos 45
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5 m/s
At the top of the arrow’s flight:
v = vx = vi (cos q ) = (96
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0°) = 68
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∆x = 27
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0°
2

ay = −g = −9
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81 m/s2)(27
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0°)(cos 50
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6 m/s
4
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0 m

Using the derivation shown in problem 1,

q = 45
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81 m/s

a
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81 m/s2)(44
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0°)(cos 45
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8 m/s

Section Two — Problem Workbook Solutions

II Ch
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At maximum height, vy, f = 0 m/s
vy, f 2 = vy, i2 + 2ay ∆ymax = 0
−(20
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0°)2
−vy, i2 −vi 2(sin q)2
ymax =  =  =  = 11
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81 m/s2)
2ay
2ay
The brick’s maximum height is 11
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−(20
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ymax = y, i =  = 22
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81 m/s2)
2ay
The brick’s maximum height is 22
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At maximum height, vy, f = 0 m/s
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∆x = 76
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0°

vy, f 2 = vy, i 2 + 2ay ∆ymax = 0
2

ay = −g = −9
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5 m)(tan 12
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07 m
4
6
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82 m/s
vi,ball = 2vrunner

In x-direction,
vi,ball (cos q ) = 2vrunner (cos q) = vrunner
2(cos q ) = 1
q = cos−1΂2΃ = 60°

7
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42 m/s
q = 55
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40 s
ay = −g = −9
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40 s
∆t1 =  = 0
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42 m/s)(sin 55
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700 s) + 2(−9
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700 s)2
∆y = 4
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40 m = 2
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44 m high
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42 m/s) (cos 55
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40 s) = 6
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vi = 2
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16 s
ay = −g = −9
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2 m/s) (cos 21°)(0
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33 m
∆t
Maximum height is reached in a time interval of 
2
2
∆t 1 ∆t
∆ymax = vi (sin q)  + 2ay 
2
2
∆ymax

΂΃ ΂΃
0
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2 m/s)(sin 21°)  +
΂2΃
−2

−2

∆ymax = 6
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1 × 10

1

2

m = 3
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2 cm

The flea’s maximum height is 3
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II Ch
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16 s
(−9
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All rights reserved

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