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NAME ______________________________________ DATE _______________ CLASS ____________________
Two-Dimensional Motion and Vectors
Problem D
PROJECTILES LAUNCHED HORIZONTALLY
PROBLEM
A movie director is shooting a scene that involves dropping s stunt
dummy out of an airplane and into a swimming pool. The plane is 10.0 m
above the ground, traveling at a velocity of 22.5 m/s in the positive x direction. The director wants to know where in the plane’s path the dummy
should be dropped so that it will land in the pool. What is the dummy’s
horizontal displacement?
SOLUTION
1. DEFINE
2. PLAN
ay = −g = −9.81 m/s2 vx = 22.5 m/s
Given:
∆y = −10.0 m
Unknown:
∆t = ?
Diagram:
The initial velocity vector of the
y
stunt dummy only has a horizontal
component. Choose the coordinate
ay
system oriented so that the positive y
direction points upward and the positive x direction points to the right.
∆x = ?
vx
x
−10.0m
Choose the equation(s) or situation: The dummy drops with no initial vertical
velocity. Because air resistance is neglected, the dummy’s horizontal velocity remains constant.
1
∆y = 2 ay (∆t)2
∆x = vx∆t
Copyright © by Holt, Rinehart and Winston. Allrights reserved.
Rearrange the equation(s) to isolate the unknown(s):
∆t =
3. CALCULATE
a
2∆y
y
First find the time it takes for the dummy to reach the ground.
∆t =
= 1.43 s
(2)(−10.0m )
(−9.81 m/s2)
Find out how far horizontally the dummy can travel during this period of time.
∆x = vx∆t = (22.5 m/s)(1.43 s)
= 32.2 m
4. EVALUATE
The stunt dummy will have to drop from the plane when the plane is at a horizontal distance of 32.2 m from the pool. The distance is within the correct order
of magnitude, given the other values in this problem.
Problem D
23
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NAME ______________________________________ DATE _______________ CLASS ____________________
ADDITIONAL PRACTICE
1. Florence Griffith-Joyner of the United States set the women’s world
record for the 200 m run by running with an average speed of 9.37 m/s.
Suppose Griffith-Joyner wants to jump over a river. She runs horizontally from the river’s higher bank at 9.37 m/s and lands on the edge of the
opposite bank. If the difference in height between the two banks is 2.00
m, how wide is the river?
2. The longest banana split ever made was 7.320 km long (needless to say,
more than one banana was used). If an archer were to shoot an arrow
horizontally from the top of Mount Everest, which is located 8848 m
above sea level, would the arrow’s horizontal displacement be larger than
7.32 km? Assume that the arrow cannot be shot faster than 100.0 m/s,
that there is no air resistance, and that the arrow lands at sea level.
3. The longest shot on a golf tournament was made by Mike Austin in
1974. The ball went a distance of 471 m. Suppose the ball was shot horizontally off a cliff at 80.0 m/s. Calculate the height of the cliff.
5. A Snorkel fire engine is designed for putting out fires that are well above
street level. The engine has a hydraulic lift that lifts the firefighter and a
system that delivers pressurized water to the firefighter. Suppose that the
engine cannot move closer than 25 m to a building that has a fire on its
sixth floor, which is 25 m above street level. Also assume that the water
nozzle is stuck in the horizontal position (an improbable situation). If
the horizontal speed of the water emerging from the hose is 15 m/s, how
high above the street must the firefighter be lifted in order for the water
to reach the fire?
6. The longest stuffed toy ever manufactured is a 420 m snake made by
Norwegian children. Suppose a projectile is thrown horizontally from a
height half as long as the snake and the projectile’s horizontal displacement is as long as the snake. What would be the projectile’s initial speed?
7. Libyan basketball player Suleiman Nashnush was the tallest basketball
player ever. His height was 2.45 m. Suppose Nashnush throws a basketball horizontally from a level equal to the top of his head. If the speed of
the basketball is 12.0 m/s when it lands, what was the ball’s initial speed?
(Hint: Consider the components of final velocity.)
8. The world’s largest flowerpot is 1.95 m high. If you were to jump horizontally from the top edge of this flowerpot at a speed of 3.0 m/s, what
would your landing velocity be?
24
Holt Physics Problem Workbook
Copyright © by Holt, Rinehart and Winston. Allrights reserved.
4. Recall Elmer Trett, who in 1994 reached a speed of 372 km/h on his motorcycle. Suppose Trett drives off a horizontal ramp at this speed and
lands a horizontal distance of 40.0 m away from the edge of the ramp.
What is the height of the ramp? Neglect air resistance.
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Givens
4. v = 925 km/h
Solutions
d1 = v∆t1 = (925 km/h)(103 m/km)(1.50 h) = 1.39 × 106 m
∆t1 = 1.50 h
d2 = v∆t2 = (925 km/h)(103 m/km)(2.00 h) = 1.85 × 106 m
∆t2 = 2.00 h
∆x1 = d1 = 1.39 × 106 m
q 2 = 135°
∆y1 = 0 m
∆x2 = d2 (cos q 2 ) = (1.85 × 106 m)(cos 135°) = −1.31 × 106 m
∆y2 = d2 (sin q 2 ) = (1.85 × 106 m)(sin 135°) = 1.31 × 106 m
∆xtot = ∆x1 + ∆x2 = 1.39 × 106 m + (− 1.31 × 106 m) = 0.08 × 106 m
∆ytot = ∆y1 + ∆y2 = 0 m + 1.31 × 106 m = 1.31 × 106 m
d = (∆
xtot
m)2
+(1.
m)2
)2+(∆
ytot)2 = (0
.0
8×106
31
×106
2 2
d = 6×
09
m2+
01
m
012m
1
1.7
2×1
= 1.
73
×1
2
d = 1.32 × 106 m = 1.32 × 103 km
∆ytot
1.31 × 106 m
q = tan−1 
= tan−1 
= 86.5° = 90.0° − 3.5°
∆xtot
0.08 × 106 m
II
q = 3.5° east of north
5. v = 57.2 km/h
d1 = v∆t1 = (57.2 km/h)(2.50 h) = 143 km
∆t1 = 2.50 h
d2 = v∆t2 = (57.2 km/h)(1.50 h) = 85.8 km
∆t2 = 1.50 h
∆tot = d1 + d2(cos q2) = 143 km + (85.8 km)(cos 30.0°) = 143 km + 74.3 km = 217 km
θ2 = 30.0°
∆ytot = d2(sin q2) = (85.8 km)(sin 30.0°) = 42.9 km
xtot
)2
+(∆
ytot
)2 = (2
)2
+(42
d = (∆
17
km
.9
km
)
d = 4.
04km
03km
04km
71
×1
2+1.8
4×1
2 = 4.
89
×1
2
Copyright © by Holt, Rinehart and Winston. All rights reserved.
d = 221 km
42.9 km
∆ytot
q = tan−1  = tan−1  = 11.2° north of east
217 km
∆xtot
Additional Practice D
1. vx = 9.37 m/s
∆y = −2.00 m
ay = −g = −9.81 m/s2
∆t =
2∆y ∆ x
 = 

v
a
x
y
∆x = vx
2∆y
(2)(−2.00 m)
 = (9.37 m/s) 
 = 5.98 m

a
(−9.81
m/s )
2
y
The river is 5.98 m wide.
2. ∆x = 7.32 km
∆y = −8848 m
ay = −g = −9.81 m/s2
∆t =
2∆y ∆ x
 = 

v
a
y
vx =
x
2
(−9.81 m/s )
 ∆x =   (7.32 × 10 m) =

2∆y
(2)(−8 848 m)
ay
3
172 m/s
No. The arrow must have a horizontal speed of 172 m/s, which is much greater than
100 m/s.
Section Two — Problem Workbook Solutions
II Ch. 3–5
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Givens
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Solutions
3. ∆x = 471 m
vi = 80.0 m/s
ay = −g = −9.81 m/s2
∆x
∆t = 
vx
ay (∆x)2 (−9.81 m/s2)(471 m)2
1
∆y = ay (∆t)2 = 
= 
= −1.70 × 102 m
2
2vx2
(2)(80.0 m/s)2
The cliff is 1.70 × 102 m high.
4. vx = 372 km/h
∆x = 40.0 m
ay = −g = −9.81 m/s2
∆x
∆t = 
vx
ay (∆x)2
1
(−9.81 m/s2)(40.0 m)2
∆y = ay (∆t)2 = 
2 = 
2
2
2vx
1 h 103 m
(2) (372 km/h)  
3600 s 1 km
∆y = −0.735 m
The ramp is 0.735 m above the ground.
II
5. ∆x = 25 m
vx = 15 m/s
ay = −g = −9.81 m/s2
h = 25 m
∆x
∆t = 
vx
ay (∆x)2 (−9.81 m/s2)(25 m)2
1
∆y = ay (∆t)2 = 
= 
(2)(15 m/s)2
2
2vx2
∆y = h − h′ = −14 m
h′ = h − ∆y = 25 m − (−14 m)
= 39 m
l = 420 m
−l
∆y = 
2
∆t =
vx =
∆x = l
2∆y
∆x
y
x
 = 

a
v
(−9.81 m/s )
 ∆x = 


(420 m) =
2∆y
(2)(−2 10m)
2
ay
64 m/s
ay = −g = −9.81 m/s2
7. ∆y = −2.45 m
v = 12.0 m/s
ay = −g = −9.81 m/s2
vy 2 = 2ay ∆y
v 2 = vx2 + vy2 = vx2 + 2ay ∆y
vx = v 2 − 2a
m
/s)2 − (2)(−9.8
m/s2)(−2.45
m)
1
y ∆y = (12.0
vx = 14
s2
−48.
s2
4m
2/
1m
2/
= 96
m2/
s2
vx = 9.8 m/s
II Ch. 3–6
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
6.
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Givens
Solutions
vy 2 = 2ay ∆y
8. ∆y = −1.95 m
vx = 3.0 m/s
ay = −g = −9.81 m/s2
v = vx 2+
vx2 + 2a
vy 2 = y ∆y
v = (3.0
m
/s)2 + (2)(−9.8
m/s2)(−1.95
m)
1
v = 9.
m2/s2 = 6.88 m/s
0m
2/s2+38.
3m
2/s2 = 47
.3
−1
q = tan
vy
2ay ∆y
(2)(−9
m/
s2)(−1.95
m)
.81
 = tan−1  = tan−1 
vx
vx
3.0 m/s
q = 64° below the horizontal
Additional Practice E
1. ∆x = 201.24 m
q = 35.0°
1
1
∆y = vi (sin q)∆t + ay (∆t)2 = vi (sin q) + ay ∆t = 0
2
2
ay = −g = −9.81 m/s2
∆x = vi (cos q)∆t
∆x
∆t = 
vi(cos q)
II
∆x
1
vi (sin q) = − ay 
2 vi(cosq)
−(−9.81 m/s )(201.24 m)
−a ∆x

 =
v =

(2)(sin 35.0°)(cos 35.0°)
2(sin q)(cos q)
2
y
i
vi = 45.8 m/s
2. ∆x = 9.50 × 102 m
Using the derivation shown in problem 1,
q = 45.0°
vi =
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ay = −g = −9.81 m/s2
−a ∆x
y =
2(sin q)(cos q)
−(−9.81 m/s2)(9.50 × 102 m)

(2)(sin 45.0°)(cos 45.0°)
vi = 96.5 m/s
At the top of the arrow’s flight:
v = vx = vi (cos q ) = (96.5 m/s)(cos 45.0°) = 68.2 m/s
3. ∆x = 27.5 m
Using the derivation shown in problem 1,
q = 50.0°
2
ay = −g = −9.81 m/s
vi =
−a ∆x
y =
2(sin q)(cos q)
−(−9.81 m/s2)(27.5 m)

(2)(sin 50.0°)(cos 50.0°)
vi = 16.6 m/s
4. ∆x = 44.0 m
Using the derivation shown in problem 1,
q = 45.0°
2
ay = −g = −9.81 m/s
a. vi =
−a ∆x
y =
2(sin q)(cos q)
−(−9.81 m/s2)(44.0 m)

(2)(sin 45.0°)(cos 45.0°)
vi = 20.8 m/s
Section Two — Problem Workbook Solutions
II Ch. 3–7