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NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 3E
PROJECTILES LAUNCHED AT AN ANGLE
PROBLEM
A flying fish leaps out of the water with a speed of 15.3 m/s. Normally
these fish use winglike fins to glide about 40 m before reentering the
ocean, but in this case the fish fails to use its “wings” and so only travels
horizontally about 17.5 m. At what angle with respect to the water’s surface does the fish leave the water? Use the trigonometric identity
2(sinq )(cosq ) = sin (2q ) to solve for q .
SOLUTION
1. DEFINE
2. PLAN
Given:
vi = 15.3 m/s
∆x = 17.5 m
g = 9.81 m/s2
Unknown:
q =?
Diagram:
vi
θ
∆x = 17.5 m
Choose the equation(s) or situation: The horizontal component of the fish’s
initial velocity, vx , is equal to the horizontal displacement divided by the time of
the jump.
∆x
vx = vi (cos q) = ∆t
The vertical displacement of the fish is given by the equation for falling bodies,
with the vertical component of the initial velocity, vy , used.
1
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆y = vy ∆t − 2 g∆t 2
Because the fish lands at the same vertical position from which it started, ∆y = 0.
∆y = 0
1
vy = vi(sin q) = 2 g∆t
Rearrange the equation(s) to isolate the unknowns: Substitute for ∆t using the
equation for horizontal velocity.
∆x
∆t = vi (cos q)
∆x
1
vi (sin q) = 2 g vi (cos q)
g∆x
(sin q)(cos q) = 2
2vi
Using the trigonometric identity allows a solution for q to be found.
1
(sin q)(cos q) = 2 [sin (2q )]
g∆x
sin (2q ) = vi2
Problem 3E
Ch. 3–13
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g∆x
sin−1 
vi 2
q=
2
(9.81 m/s2)(17.5 m)
sin−1 
(15.3 m/s)2
q=
2
3. CALCULATE
= 23.6° above the horizontal
4. EVALUATE
Substituting the value for q into the original equations and solving for ∆t produces a time of 1.25 s for both, thus confirming the result for q .
ADDITIONAL PRACTICE
1. A baseball is thrown with an initial speed of 15.0 m/s. If the ball’s horizontal displacement is 17.6 m, at what angle with respect to the ground is
the ball pitched? Use the trigonometric identity 2(sin q )(cos q ) = sin
(2q ) to solve for q .
2. A football is kicked so that its initial speed is 23.1 m/s. If the football
reaches a maximum height of 16.9 m, at what angle with respect to the
ground is the ball kicked?
4. The small jumping spiders make up for their size by their ability to leap
relatively large distances. Some can jump fifty times the length of their
bodies. Suppose a jumping spider leaps a horizontal distance of 18.5 cm
with an initial speed of about 141 cm/s. At what angle above the horizontal would a spider with this speed have to leap in order to travel a range
of 18.5 cm? Use the trigonometric identity 2(sin q )(cos q ) = sin (2q ) to
solve for q .
5. Olympic platform divers jump from a diving board that is 10.0 m above
the water. Suppose a diver jumps from the board with an initial speed of
6.03 m/s. The diver reaches a maximum height of 11.7 m above the
water, and lands in the water at a horizontal distance of 3.62 m from the
end of the board. At what angle with respect to the board does the diver
leave the board?
6. A ball is thrown from a roof with a speed of 10.0 m/s and an angle of
37.0° with respect to the horizontal. What are the vertical and horizontal
components of the ball’s displacement 2.5 s after it is thrown?
7. A downed pilot fires a flare from a flare gun. The flare has an initial
speed of 250 m/s and is fired at an angle of 35° to the ground. How long
does it take for the flare to reach its maximum altitude?
Ch. 3–14
Holt Physics Problem Bank
Copyright © by Holt, Rinehart and Winston. All rights reserved.
3. Jackie Joyner-Kersee’s record long jump is 7.49 m. Suppose she ran
9.50 m/s to jump this horizontal distance. At what angle above the horizontal did she jump? Use the trigonometric identity 2(sin q )(cos q ) =
sin (2q ) to solve for q .
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8. In the sport of ski jumping, a skier travels down the slope of a hill until
he or she reaches the takeoff. The takeoff is slanted slightly below the
horizontal, so that the skier is able to travel in the air just above the
ground. Suppose a skier leaves the takeoff and lands 73.0 m horizontally beyond the takeoff and −52.8 m below the takeoff. If the takeoff
angle is −8.00° below the horizontal, what is the skier’s initial speed?
9. A shingle slides down a roof having a 30.0° pitch and falls off with a
speed of 2.0 m/s. How long will it take to hit the ground 45 m below?
Copyright © by Holt, Rinehart and Winston. All rights reserved.
10. A hole at a miniature golf course requires the ball to roll up a ramp, fly
over a small stream, and then land on the green beyond the stream.
The stream is 0.46 m wide, and the cup is 4.00 m beyond the stream’s
edge. The ramp makes an angle of 41.0° with the horizontal, and its
upper edge is 0.35 m above the green. What must the ball’s initial speed
be in order for the ball to fly over the water and land directly in the
cup?
Problem 3E
Ch. 3–15
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Givens
Solutions
9. vx = 1.50 m/s
∆y = −2.50 × 102 m
2
g = 9.81 m/s
vy,f 2 = −2g∆y + vy,i2
vy,i = 0 m/s, so
vy,f = vy = −
m/s
m)
2g
∆
y = −(2
)(
9.
81
2)(−
2.
50
×102
vy = 70.0 m/s
2
m/s
m2/s2
+4.9
m2/s2
v = vx2+
v
.5
0m
/s
)2+(70
.0
)2 = 2.
25
0×103
y = (1
= 4.
03
m2/
s2
90
×1
v = 70.0 m/s
v
1.50 m/s
q = tan−1 x = tan−1 
vy
70.0 m/s
q = 1.23° from the vertical
10. vx = 85.3 m/s
∆t =
x
∆y = −1.50 m
g = 9.81 m/s2
2∆y ∆x
−g = v
∆x = vx
(2)(−1.50 m)
2∆y
−g = (85.3 m/s) 
−9.81
m/s = 47.2 m
2
range of arrow = 47.2 m
Additional Practice 3E
1. vi = 15.0 m/s
∆x = 17.6 m
∆x
∆t = 
vi(cos q)
1
∆y = vi(sin q)∆t − 2g∆t 2 = 0
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆x
1
1
vi(sin q) = 2g∆t = 2g 
vi(cos q)
g∆x
2(sin q)(cos q) = 
vi 2
Using the identity 2 (sin q)(cos q) = sin (2 q),
g∆x
sin (2 q) = 
vi 2
g∆x
(9.81 m/s2)(17.6 m)
sin−1 
sin − 
2
vi
(15.0 m/s)2
q = 
= 
2
2
q = 25.1°
2. vi = 23.1 m/s
vy,f 2 − vy,i2 = −2g∆y
∆ymax = 16.9 m
At maximum height, vy,f
g = 9.81 m/s2
vy,i = vi (sin q) = 2g
ymax
∆
−1
q = sin
2g
∆
y
m
ax
(2
m/s
)(
9.
81
2)(1
6.
9m
)

= sin 
23.1 m/s
v
−1
i
q = 52.0°
V
Section Five—Problem Bank
V Ch. 3–11
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Givens
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Solutions
3. ∆x = 7.49 m
vi = 9.50 m/s
g = 9.81 m/s2
Using the form of the equation derived in problem 1,
g∆x
2(sin q)(cos q) = sin(2 q) = 
vi2
(9.81 m/s2)(7.49 m)
g∆x


2
(9.50 m/s2
−1 vi
−1 
q = sin  = sin
2
2
q = 27.3°
4. vi = 141 cm/s
∆x = 18.5 cm
g = 9.81 m/s2
Using the form of the equation derived in problem 1,
g∆x
2(sin q)(cos q) = sin(2 q) = 
vi 2
g∆x

vi2
q = sin−1
2
(9.81 m/s2)(18.5 × 10−2 m)
sin−1 
(141 × 10−2 m/s)2
2
q = 33.0°
5. vi = 6.03 m/s
vy,f 2 − vy,i 2 = −2g∆y
hi = 10.0 m
At maximum height, vy,f = 0.
hmax = hf = 11.7 m
vy,i = vi(sin q) = 2g
ym
∆
ax
∆x = 3.62 m
For the diver, hf is the maximum height above the diving board.
g = 9.81 m/s2
∆y = hf − hi
q = sin
2g
(h
f−hi)
(2
)(
9.
81
)m
/s
2)(1
1.
7m
−10.
0m
)

= sin 
v
6.03 m/s
−1
−1
i
6. vi = 10.0 m/s
∆x = vi(cos q)∆t = (10.0 m/s)(cos 37.0°)(2.5 s)
q = 37.0°
∆x = 2.0 101 m
∆t = 2.5 s
∆y = vi(sin q)∆t − 2g∆t 2 = (10.0 m/s)(sin 37.0°)(2.5 s) − 2(9.81 m/s2)(2.5 s)2
1
g = 9.81 m/s2
= 15 m − 31 m
∆y = −16 m
7. vi = 250 m/s
At the maximum height
q = 35°
vy,f = vy,i − g∆t = 0
2
g = 9.81 m/s
vy,i = vi(sin q) = g∆t
vi(sin q) (250 m/s)(sin 35°)
 = 
∆t = 
g
9.81 m/s2
∆t = 15 s
V
V Ch. 3–12
Holt Physics Solution Manual
1
Copyright © by Holt, Rinehart and Winston. All rights reserved.
q = 73.3°
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Givens
8. ∆x = 73.0 m
∆y = −52.8 m
q = −8.00°
g = 9.81 m/s2
Print
Solutions
∆x
∆t = 
vi(cos q)
2
∆x
∆x
1
1
∆y = vi(sin q)∆t − 2g∆t 2 = vi (sin q)  − 2g 
vi (cos q)
vi(cos q)
g∆x2
∆y = ∆x(tanq) − 
2 2
2 vi (cos q)
g∆x2

2 2 = ∆x(tan q) − ∆y
2 vi (cos q)
g∆x2
2vi2(cos q)2 = 
[∆x(tan q) − ∆y]
vi =
vi =
g∆x2

2
2(cos q) [∆x(tan q) − ∆y]
(9.81 m/s2)(73.0 m)2

(2)[cos(−8.00°)]2 [(73.0 m)(tan[−8.00°]) − (−52.8 m)]
(9.81 m/s2)(73.0 m)2

(2)[cos(−8.00°)]2 (−10.3 m + 52.8 m)
vi =
vi =

(2)[−cos(–8.00°)] (42.5 m)
(9.81 m/s2)(73.0 m)2
2
vi = 25.0 m/s
9. q = −30.0°
vi = 2.0 m/s
∆y = −45 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
g = 9.81 m/s2
1
∆y = vi (sin q)∆t − 2g∆t 2
2 ∆t 2 − [vi(sin q)]∆t + ∆y = 0
g
Solving for ∆t using the quadratic equation,
vi(sin q) ± [−vi(sin q)]2 − 42(∆y)
∆t = 
g
22
g
(2.0 m/s)[sin(−30.0°)] ± [(
]2
−(2)
m)
−2.
0m
/s
)[
si
n(−
30
.0
°)
(9
.8
1m
/s
2)(−
45
∆t = 
2
9.81 m/s
−1.0 m/s ± 1.
s2
+8.8
02
m2/
s2 −1.0 m/s ± 8.
02
m2/
s2
0m
2/
×1
8×1

∆t = 
=
9.81 m/s2
9.81 m/s2
−1.0 m/s ± 3.0 × 101 m/s
∆t = 
9.81 m/s2
∆t must be positive, so the positive root must be chosen.
29 m/s
∆t = 2 = 3.0 s
9.81 m/s
V
Section Five—Problem Bank
V Ch. 3–13
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Givens
Solutions
10. ∆x1 = 0.46 m
∆xtot = ∆x1 + ∆x2
∆x
∆t = 
vi(cos q)
∆x2 = 4.00 m
∆xi + ∆x2 1 ∆x1 + ∆x2
1
 − 2g 
∆y = vi(sin q)∆t − 2g∆t 2 = vi(sin q) 
vi(cos q)
vi(cos q)
∆y = − 0.35 m
g(∆x1 + ∆x2 )2

∆y = (∆x1 + ∆x2)(tan q) − 
2 vi2(cos q)2
g = 9.81 m/s2
vi =
vi =
vi =
vi =
2
q = 41.0°
g(∆x1 + ∆x2)2

2
2(cos q) [(∆x1 + ∆x2)(tan q) − ∆y]
(9.81 m/s2)(0.46 m + 4.00 m)2

2
(2)(cos 41.0°) [(0.46 m + 4.00 m)(tan 41.0°) − (−0.35 m)]
(9.81 m/s2)(4.46 m)2

(2)(cos 41.0°)2(3.88 m + 0.35 m)
(9.81 m/s2)(4.46 m)2

(2)(cos 41.0°)2 (4.23 m)
vi = 6.36 m/s
Additional Practice 3F
1. vbw = 58.0 km/h, forward
= +58.0 km/h
vwe = 55.0 km/h, backward
= −55.0 km/h
vbe = vbw + vwe = + 58.0 km/h + (−55.0 km/h) = +3.0 km/h
∆x 1.4 km
∆t =  = 
vbe 3.0 km/h
∆t = 0.47 h = 28 min
∆x = 1.4 km
vmw = 4.20 m/s, west
2
∆x = 8.50 × 10 m
vme = vmw + vwe = 4.20 m/s + 1.50 m/s = 5.70 m/s, west
time of travel with walkway:
∆x 8.50 × 102 m
∆t1 =  =  = 149 s
vme
5.70 m/s
time of travel without walkway:
∆x 8.50 × 102 m
∆t2 =  =  = 202 s
vmw
4.20 m/s
time saved = ∆t2 − ∆t1 = 202 s − 149 s = 53 s
3. v1e = 286 km/h, forward
v12 + v2e = v1e
v2e = 252 km/h, forward
v12 = v1e − v2e
∆x = 0.750 km
v12 = v1e − v2e = 286 km/h − 252 km/h = 34 km/h
∆x 0.750 km
∆t =  −  = 2.2 × 10−2 h
v12 34 km/h
3600 s
∆t = (2.2 × 10−2 h)  = 79 s
1h
V
V Ch. 3–14
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
2. vwe = 1.50 m/s, west