8) A soccer ball is kicked horizontally off a 22.0-meter high hill and lands a
distance of 35.0 meters from the edge of the hill. Determine the initial horizontal
velocity of the soccer ball.
Horizontal Info:
Vertical Info:
x = 35.0 m
y = -22.0 m
vix = ???
viy = 0 m/s
ax = 0 m/s/s
ay = -9.8 m/s/s
Use y = viy • t + 0.5 • ay • t2 to solve for time; the time of flight is 2.12 seconds.
Now use x = vix • t + 0.5 • ax • t2 to solve for vix
Note that ax is 0 m/s/s so the last term on the right side of the equation cancels. By substituting 35.0 m
for x and 2.12 s for t, the vix can be found to be 16.5 m/s.
9) A long jumper leaves the ground with an initial velocity of 12 m/s at an angle of 28degrees above the horizontal. Determine the time of flight, the horizontal distance, and
the peak height of the long-jumper.
As a first step, use trigonometric function to determine the initial velocity components:
Horiz: vix =(12 m/s) • cos (28 deg) =10.6 m/s
Vert: viy =(12 m/s) • sin (28 deg) = 5.6 m/s
Then set up an x-y table, listing the known information.
Horizontal Info:
x = ???
Vertical Info:
y = 0.0 m (total); ypeak = ???
viy = 5.6 m/s
vix = 10.6 m/s
vy-peak = 0 m/s
vfy = -5.6 m/s
ax = 0 m/s/s
ay = -9.8 m/s/s
Use vfy = viy + ay • ttotal and vertical info to solve for time; the time of flight is 1.1 seconds (rounded
from 1.1497 s).
Now use x = vix • t + 0.5 • ax • t2 to solve for x. Note that ax is 0 m/s/s so the last term on the right side
of the equation cancels. By substituting 10.6 m/s for vix and 1.1 s for t, the x can be found to be 12.2 m
(rounded from 12.1817 m)
C) 1.6 meters
10) A football is kicked with an initial velocity of 25 m/s at an angle of 45-degrees with the
horizontal. Determine the time of flight, the horizontal displacement, and the peak height
of the football.
The solution of any non-horizontally launched projectile problem (in which vi and Theta are
given) should begin by first resolving the initial velocity into horizontal and vertical
components using the trigonometric functions discussed above. Thus,
Horizontal Component
vix = vi•cos(Theta)
Vertical Component
viy = vi•sin(Theta)
vix = 25 m/s•cos(45 deg)
viy = 25 m/s•sin(45 deg)
vix = 17.7 m/s
viy = 17.7 m/s
In this case, it happens that the vix and the viy values are the same as will always be the case when the
angle is 45-degrees.
The solution continues by declaring the values of the known information in terms of the symbols of the
kinematic equations - x, y, vix, viy, ax, ay, and t. In this case, the following information is either
explicitly given or implied in the problem statement:
Horizontal Information
Vertical Information
x = ???
y = ???
vix = 17.7 m/s
viy = 17.7 m/s
vfx = 17.7 m/s
vfy = -17.7 m/s
ax = 0 m/s/s
ay = -9.8 m/s/s
As indicated in the table, the final x-velocity (vfx) is the same as the initial x-velocity (vix). This is due
to the fact that the horizontal velocity of a projectile is constant; there is no horizontal acceleration.
The table also indicates that the final y-velocity (vfy) has the same magnitude and the opposite
direction as the initial y-velocity (viy). This is due to the symmetrical nature of a projectile's trajectory.
The unknown quantities are the horizontal displacement, the time of flight, and the height of the
football at its peak. The solution of the problem now requires the selection of an appropriate strategy
for using the kinematic equations and the known information to solve for the unknown quantities.
There are a variety of possible strategies for solving the problem. An organized listing of known
quantities in two columns of a table provides clues for the selection of a useful strategy.
From the vertical information in the table above and the second equation listed among the vertical
kinematic equations (vfy = viy + ay*t), it becomes obvious that the time of flight of the projectile can be
determined. By substitution of known values, the equation takes the form of
-17.7 m/s = 17.7 m/s + (-9.8 m/s/s)•t
The physics problem now takes the form of an algebra problem. By subtracting 17.7 m/s from each
side of the equation, the equation becomes
-35.4 m/s = (-9.8 m/s/s)•t
If both sides of the equation are divided by -9.8 m/s/s, the equation becomes
3.61 s = t
(rounded from 3.6077 s)
The total time of flight of the football is 3.61 seconds.
With the time determined, information in the table and the horizontal kinematic equations can be used
to determine the horizontal displacement (x) of the projectile. The first equation (x = vix•t + 0.5•ax•t2)
listed among the horizontal kinematic equations is suitable for determining x. With the equation
selected, the physics problem once more becomes transformed into an algebra problem. By
substitution of known values, the equation takes the form of
x = (17.7 m/s)•(3.6077 s) + 0.5•(0 m/s/s)•(3.6077 s)2
Since the second term on the right side of the equation reduces to 0, the equation can then be simplified
to
x = (17.7 m/s)•(3.6077 s)
Thus,
x = 63.8 m
The horizontal displacement of the projectile is 63.8 m.
Finally, the problem statement asks for the height of the projectile at is peak. This is the same as
asking, "what is the vertical displacement (y) of the projectile when it is halfway through its
trajectory?" In other words, find y when t = 1.80 seconds (one-half of the total time). To determine the
peak height of the projectile (y with t = 1.80 sec), the first equation (y = viy•t +0.5•ay•t2) listed among
the vertical kinematic equations can be used. By substitution of known values into this equation, it
takes the form of
y = (17.7 m/s)•(1.80 s) + 0.5*(-10 m/s/s)•(1.80 s)2
Using a calculator, this equation can be simplified to
y = 31.9 m + (-15.9 m)
And thus,
y = 15.9 m
The solution to the problem statement yields the following answers: the time of flight of the football is
3.61 s, the horizontal displacement of the football is 63.8 m, and the peak height of the football 15.9 m.
11) A golf ball is hit with a velocity of 24.5 m/s at 35.0 degrees above the horizontal.
Find
a) the range of the ball, and
Known:
Up will be considered positive and down will be considered negative
a = g = -9.81 m/s2
vi = 24.5 m/s @ 35.0°
y=0m
x=?
Solution:
First, determine the horizontal and vertical components of the initial velocity.
v

v


(
24
.
5
m
/
s
)(cos
35
)

20
.
0692
m
/
s
x
icos
v

v


(
24
.
5
m
/
s
)(sin
35
)

14
.
0526
m
/
s
iy
isin

Need to determine the time it took to land:
yviyt  21 at2
0m(14.0526
m/s)t  21 (9.81
m/s2)t2
0m(14.0526
/s4.905
m/s2t)t
t 0s
0m14.0526
/s4.905
m/s2t
14.0526
m/s
t
4.905
m/s2
t 2.8649
s
xvxt

x(20
.0692
m
/s)(
2.8649
s)
x59
.4963
m
The range of the ball is 59.5 meters.

b) the
maximum height of the ball.
Half the flight time will be 1.4324 seconds.
1 2
y

v
t
at
iy
2
2
2
1
y

(
14
.
0526
m
/s
)(
1
.
4324
s
)
(

9
.
81
m
/s
)(
1
.
4324
)
2
y

20
.
1289
m

10
.
0639
m
y

10
.
0650
m
The ball will reach a maximum height of 10.1 meters.

12)
A player kicks a football from ground level at 27.0 m/s at an angle of 30.0
degrees above the horizontal. Find
a) its "hang time" (time that the ball is in the air),
Known:
Up will be considered positive and down will be considered negative
a = g = -9.81 m/s2
vi = 27.0 m/s @ 30.0°
y=0m
t=?
Solution:
First, determine the horizontal and vertical components of the initial velocity.
v

v


(
27
m
/
s
)(cos
30
)

23
.
3827
m
/
s
x
icos
v

v


(
27
m
/
s
)(sin
30
)

13
.
5
m
/
s
iy
isin
y  viyt  21 at2

0m (13.5m/ s)t  21 (9.81m/ s2)t2
0m (13.5/ s4.905m/ s2t)t
t  0s
0m13.5/ s4.905m/ s2t
13.5m/ s
t
4.905m/ s2
t  2.7523s
The ball was in the air for 2.75 seconds.
distance the ball travels before it hits the ground, and
b) the
xvxt
x(23
.3727
m
/s)(
2.7523
s)
x64
.3287
m
The ball traveled 64.3 meters
c) its
maximum height.
Half the flight time will be 1.3762 seconds.
1 2
y

v
t
at
iy
2
2
2
1
y

(
13
.5
m
/s
)(
1
.3762
s
)
(

9
.81
m
/s
)(
1
.3762
)
2
y

18
.5787
m

9
.2697
m
y

9
.3090
m
The ball will reach a maximum height of 9.31 meters.
13). 
After retrieving her ball from the roof, a girl throws the ball off of the roof with a
speed of 5.6 m/s at an angle of 35 degrees above the horizontal. The ball is thrown
from a height of 11 m above the ground. Find
a) the time the ball is in the air,
Known:
Up will be considered positive and down will be considered negative
a = g = -9.81 m/s2
vi = 5.6 m/s @ 35°
y = -11 m
x=?
Solution:
First, determine the horizontal and vertical components of the initial velocity.
v

v


(
5
.
6
m
/
s
)(cos
35
)

4
.
5873
m
/
s
x
icos
v

v


(
5
.
6
m
/
s
)(sin
35
)

3
.
2120
m
/
s
iy
isin
1 2
y
v
t
at
iy
2

2 2
1

11
m

(3
.2120
m
/s
)t
(
9
.81
m
/s
)t
2
2 2
(4
.905
m
/s
)t 
(3
.2120
m
/s
)t
11
m

0
Use the quadratic formula:

t

b  b2  4ac
2a
(3.2120)  (3.2120)2  4(4.905)(11)
2(4.905)
3.2120 10.3169 215.82
9.81
3.2120 226.1369 3.212015.0378


9.81
9.81
3.212015.0378 3.212015.0378

or
9.81
9.81
18.2498 11.8258

or
9.81
9.81
 1.8603s or 1.2055s

The ball will land after 1.9 seconds.

b) the range of the ball,
Solution:
xvxt
x(4.5873
m/s)(
1.8603
s)
x8.5338
m
The ball will travel 8.5 meters away from the base of the building.

c) the
maximum height of the ball, and
Solution:
In this case (because the take-off and landing heights are not the same) the
maximum height will not occur at the half-way time. We do however, know that
the maximum height occurs when vy = 0 m/s.
Find the distance when vy = 0 m/s
2
2
vfy vi y 2
ay
(0
m
/s)2 (3
.2120
m
/s)22
(
9
.81
m
/s2)y
2
010
.3169
m
/s2(19
.62
m
/s2)y
2
10
.3169
m
/s2
y
19
.62
m
/s2
y0
.5258
m

This gives us the distance above the starting point. But since the starting point
is 11 meters off the ground we also need to take that into account.
The ball reaches a maximum height of 11.5 meters above the ground.
d) the speed of the ball as it strikes the ground.
Because vx doesn’t change we know the horizontally velocity of the ball remains
constant at 4.5873 m/s.
Need to determine the final velocity in the vertical direction.
v
v
at
f y
iy
2
v
3
.2120
m
/s(
9
.81
m
/s
)(
1
.8603
s
)
f y
v
3
.2120
m
/s
18
.2495
f y
v

15
.0375
f y
To determine the final speed we need to resolve the perpendicular vectors.

2
2
2
vf vf x vf y
2
vf (4.5873
m/s)2 (15
.0375
m/s)2
2
vf 21
.0433
m2 /s2 226
.1264
m2 /s2
2
vf 247
.1697
m2 /s2
vf  247
.1697
m2 /s2
vf 15
.7216
m/s
The ball hits the ground with a speed of 15.7 m/s.
