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NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 3E
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. The strait is only about 6.0 m
wide. Suppose an athlete wanting to jump “over the sea” leaps at an angle
of 35° with respect to the horizontal. What is the minimum initial speed
that would allow the athlete to clear the gap? Neglect air resistance.
SOLUTION
1. DEFINE
2. PLAN
Given:
∆x = 6.0 m
q = 35°
g = 9.81 m/s
Unknown:
vi = ?
Diagram:
v
θ = 35°
∆x
= 6.00 m
∆x
vx = vi cos q = 
∆t
At the midpoint of the jump, the vertical component of the athlete’s velocity, vy ,
which is the upward vertical component of the initial velocity, vi sin q, minus the
downward component of velocity due to free-fall acceleration, equals zero. The
time required for this to occur is half the time necessary for the total jump.
∆t
vy = vi sin q − g  = 0
2
g ∆t
vi sin q = 
2
Rearrange the equation(s) to isolate the unknown(s): Express ∆t in the second
equation in terms of the displacement and velocity component in the first equation.
g ∆x
vi sin q =  
2 vi cos q
g ∆x
vi 2 = 
2 sin q cos q
24
Holt Physics Problem Workbook
Copyright © by Holt, Rinehart and Winston. Allrights reserved.
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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vi =
3. CALCULATE
g ∆x

2 sinq
cos q
Substitute the values into the equation(s) and solve: Select the positive root for vi .
m
9.81 (6.0 m)
s
v = 
2
i
4. EVALUATE
(2)(sin 35°)(cos 35°)
m
= 7.9 
s
By substituting the value for vi into the original equations, you can determine the
time for the jump to be completed, which is 0.92 s. From this, the height of the
jump is found to equal 1.0 m.
ADDITIONAL PRACTICE
1. In 1993, Wayne Brian threw a spear a record distance of 201.24 m. (This
is not an official sport record because a special device was used to “elongate” Brian’s hand.) Suppose Brian threw the spear at a 35.0° angle with
respect to the horizontal. What was the initial speed of the spear?
2. April Moon set a record in flight shooting (a variety of long-distance
archery). In 1981 in Utah, she sent an arrow a horizontal distance of
9.50 × 102 m. What was the speed of the arrow at the top of the flight if the
arrow was launched at an angle of 45.0° with respect to the horizontal?
Copyright © by Holt, Rinehart and Winston. Allrights reserved.
3. In 1989 during overtime in a high school basketball game in Erie, Pennsylvania, Chris Eddy threw a basketball a distance of 27.5 m to score and
win the game. If the shot was made at a 50.0° angle above the horizontal,
what was the initial speed of the ball?
4. In 1978, Geoff Capes of the United Kingdom won a competition for
throwing 5 lb bricks; he threw one brick a distance of 44.0 m. Suppose
the brick left Capes’ hand at an angle of 45.0° with respect to the
horizontal.
a. What was the initial speed of the brick?
b. What was the maximum height reached by the brick?
c. If Capes threw the brick straight up with the speed found in (a),
what would be the maximum height the brick could achieve?
5. In 1991, Doug Danger rode a motorcycle to jump a horizontal distance
of 76.5 m. Find the maximum height of the jump if his angle with respect to the ground at the beginning of the jump was 12.0°.
6. Michael Hout of Ohio can run 110.0 meter hurdles in 18.9 s at an average speed of 5.82 m/s. What makes this interesting is that he juggles three
balls as he runs the distance. Suppose Hout throws a ball up and forward
at twice his running speed and just catches it at the same level. At what
angle, q , must the ball be thrown? (Hint: Consider horizontal displacements for Hout and the ball.)
Problem 3E
25
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7. A scared kangaroo once cleared a fence by jumping with a speed of
8.42 m/s at an angle of 55.2° with respect to the ground. If the jump
lasted 1.40 s, how high was the fence? What was the kangaroo’s horizontal displacement?
Copyright © by Holt, Rinehart and Winston. Allrights reserved.
8. Measurements made in 1910 indicate that the common flea is an impressive jumper, given its size. Assume that a flea’s initial speed is 2.2 m/s, and
that it leaps at an angle of 21° with respect to the horizontal. If the jump
lasts 0.16 s, what is the magnitude of the flea’s horizontal displacement?
How high does the flea jump?
26
Holt Physics Problem Workbook
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Givens
Solutions
8. ∆y = −1.95 m
vx = 3.0 m/s
g = 9.81 m/s2
vy 2 = −2g ∆y
v = vx 2+
vy 2 = vx2−2g∆
y
v = (3
m/s
)2
−(2)
m)
.0
(9
.8
1m
/s
2)(−
1.
95
v = 9.
m2/s2 = 6.88 m/s
0m
2/s2+38.
3m
2/s2 = 47
.3
−1
q = tan
2g
∆y
−
−
m/s
vy
(2
)(
9.
81
2)(–1
.9
5m
)
 = tan−1  = tan−1 
v
3.0
m/s
vx
x
q = 64° below the horizontal
Additional Practice 3E
1. ∆x = 201.24 m
q = 35.0°
1
1
∆y = vi (sin q) ∆t − g∆t 2 = vi (sin q) − g∆t = 0
2
2
g = 9.81 m/s2
∆x = vi (cos q)∆t
II
∆x
∆t = 
vi(cos q)
(9.81 m/s )(201.24 m)

v = 
2(singq∆)
(xcosq) = (2)(sin 35.0°)(cos 35.0°)
∆x
1
vi(sin q) = g  
2 vi (cos q)
2
i
vi = 45.8 m/s
2. ∆x = 9.50 × 102 m
q = 45.0°
Using the derivation shown in problem 1,
vi =
Copyright © by Holt, Rinehart and Winston. All rights reserved.
g = 9.81 m/s2
x
(9.81 m/s )(9.50 × 10 m)


2(sinqg∆)(
cos q ) = (2)(sin 45.0°)(cos 45.0°)
2
2
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
g = 9.81 m/s
vi =
g ∆x
 =
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
q = 45.0°
g = 9.81 m/s2
Using the derivation shown in problem 1,
a. vi =
g ∆x
 =
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
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Solutions
b. At maximum height, vy, f = 0 m/s
vy, f2 = vy, i2 − 2g∆y = 0
vy, i2 vi 2(sin q)2
(20.8 m/s)2(sin 45.0°)2
 = 
∆ymax =  = 
= 11.0 m
2g
(2)(9.81 m/s2)
2g
The brick’s maximum height is 11.0 m.
(20.8 m/s)2
v2
c. ∆ymax = i = 2 = 22.1 m
2g (2)(9.81 m/s
The brick’s maximum height is 22.1 m.
5. ∆x = 76.5 m
q = 12.0°
At maximum height, vy, f = 0 m/s.
vy, f 2 = vy, i 2 − 2g∆y = 0
2
g = 9.81 m/s
vy, i2 vi 2(sin q)2

∆ymax =  = 
2g
2g
Using the derivation for vi 2 from problem 1,
g∆x
(sin q)2 ∆x (sin q) ∆x (tan q)
∆ymax =   =  = 
2(sin q)(cos q)
2g
4(cos q)
4
II
(76.5 m)(tan 12.0°)
∆ymax =  = 4.07 m
4
6. vrunner = 5.82 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. vi = 8.42 m/s
q = 55.2°
∆t = 1.40 s
g = 9.81 m/s2
For first half of jump,
1.40 s
∆t1 =  = 0.700 s
2
1
1
∆y = vi (sin q)∆t1 − 2g∆t12 = (8.42 m/s)(sin 55.2°)(0.700 s) − 2(9.81 m/s2)(0.700 s)2
∆y = 4.84 m − 2.40 m = 2.44 m
The fence is 2.44 m high.
∆x = vi (cos q)∆t
∆x = (8.42 m/s) (cos 55.2°)(1.40 s) = 6.73 m
8. vi = 2.2 m/s
q = 21°
∆t = 0.16 s
g = 9.81 m/s2
∆x = vi (cos q)∆t = (2.2 m/s) (cos 21°)(0.16 s) = 0.33 m
∆t
Maximum height is reached in a time interval of 
2
2
∆t 1 ∆t
∆ymax = vi (sin q)  − 2g 
2
2
∆ymax
0.16 s
= (2.2 m/s)(sin 21°)  −
2
−2
−2
∆ymax = 6.3 × 10 m − 3.1 × 10
1

2
m = 3.2 × 10 m = 3.2 cm
The flea’s maximum height is 3.2 cm.
II Ch. 3–8
Holt Physics Solution Manual
−2
2
0.16 s
(9.81 m/s2) 
2
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1