Quiz
A boy on a small hill with a slingshot is below a boy in a tree, and so aims upward, directly at the boy in the tree. Does the boy in the tree make the right move by letting go at the moment the water balloon is shot? Explain.
Both water balloon and boy experience the same acceleration since gravity causes all objects to accelerate at the same rate regardless of their mass.
Since both balloon and boy experience the same acceleration each will fall equal amounts below their gravity-free path.

KINEMATICS LESSON 7 PART 2
Non-Horizontally Launched Projectile Problems
What is the difference between horizontally and non horizontally launched projectile problems?
How do we solve non-horizontally launched projectile problems?
First we determine the vertical and horizontal components of the initial velocity of the projectile.
Second, we use the same equations as horizontal projectile problems, but now v iy
does not = 0 but has an actual value.
Δd y
= v iy
+ ½ at 2 , where Δd = d f
- d i v fy
= v iy
+ a y t v fy
2 = v iy
2 + 2a y
Δd, where Δd = d f
- d i
Δd x
= v ix
• t
Example
A football is kicked with an initial velocity of 25 m/s at an angle of 45degrees with the horizontal. Determine the time of flight, the horizontal displacement, and the peak height of the football.
FIRST, draw a diagram of the problem
SECOND, Choose an origin and an xy coordinate system
Since the football is kicked from the ground, its x-y coordinate system is
(0,0)

THIRD, resolve the initial velocity into horizontal and vertical components using the trig functions. Thus,
Horizontal Component v ix
= v i
•cos(Theta) v ix
= 25 m/s•cos(45 deg) v ix
= 17.7 m/s v iy
Vertical Component v iy
= v i
•sin(Theta)
= 25 m/s•sin(45 deg) v iy
= 17.7 m/s
FOURTH, Analyze the horizontal motion and the vertical motion.
Horizontal Information d fx
= ??? d ix
= 0 v ix
= 17.7 m/s v fx
= 17.7 m/s a x
= 0 m/s/s
Vertical Information d fy
= ??? d iy
= 0 v iy
= 17.7 m/s v fy
= -17.7 m/s a y
= -9.8 m/s/s
As indicated in the table, the final x-velocity Note v fx
= v ix
. This is due to the fact that the since there is no horizontal acceleration.
Also, v fy
has the same magnitude and the opposite direction as v iy
. 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.
FIFTH, devise an appropriate strategy for using the kinematic equations and the known information to solve for the unknown quantities.
From the vertical information it becomes obvious that the time of flight of the projectile can be determined using the equation (v fy
= v iy
+ a y
*t),
-17.7 m/s = 17.7 m/s + (-9.8 m/s/s)•t
-35.4 m/s = (-9.8 m/s/s)•t
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
(d x
) of the projectile using the equation Δd x
= v ix t, d xf
= (17.7 m/s)•(3.6077 s) + 0.5•(0 m/s/s)•(3.6077 s) 2 d xf
= (17.7 m/s)•(3.6077 s) d xf
= 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 (d 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). Using the equation, Δd y
= v iy t+ ½ at 2 , where Δd = d f
- d i
, we can find the peak height d yf
- 0 = (17.7 m/s)•(1.80 s) + 0.5*(-10 m/s/s)•(1.80 s) 2 d yf
= 31.9 m + (-15.9 m) d yf
= 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.
(Note that in all calculations performed above, unrounded numbers were used. The numbers reported in the preliminary steps and in the final answer were the rounded form of the actual unrounded values.)
ANIMATION
Satellite Motion – free fall

HOMEWORK KINEMATICS LESSON 7 PART 2
1. A long jumper leaves the ground with an initial velocity of 12 m/s at an angle of 28-degrees above the horizontal. Determine the time of flight, the horizontal distance, and the peak height of the long-jumper.
2. A football is kicked at an angle of 37.0 degrees above the horizontal with a velocity of 20.0 m/s. Calculate (a) the maximum height, (b) the time of travel before the football hits the ground, (c) how far away it hits the ground, (d) the velocity vector at the maximum height, and (e) the acceleration vector at maximum height . Assume the ball leaves the foot at ground level.
3. A child sits upright in a wagon which is moving to the right at constant speed. The child ext3ends her hand and throws an apple straight upward
(from her point of view), while the wagon continues to travel forward at constant speed. If air resistance is neglected, will the apple land (a) behind the wagon, (b) in the wagon, or (c) in front of the wagon? Explain.
4. A projectile is fired with an initial speed of 75.2 m/s at an angle of 34.5
o above the horizontal on a long flat firing range. Determine (a) the maximum height reached by the projectile, (b) the total time in the air, (c) the total horizontal distance covered, and (d) the velocity of the projectile
1.50 s after firing and its angle above the horizontal. Hint, find the vertical and horizontal velocities after 1.50 s and then use Pythagoras.
5. The pilot of an airplane traveling 160 km/h wants to drop supplies to flood victims isolated on a patch of land 160 m below. The supplies should be dropped how many seconds before the plane is directly overhead?
ANSWER KEY
1.
1.1 s, 12 m, 1.6 m
2.
7.35 m, 2.45s, 39.2 m/s, 16.0 m/s, 9.8 m/s
2
downward.
3.
no key
4.
92.6 m, 8.69 s, 5.40 x 10
2
m, 68.2 m/s, 22.2
o
5.
5.7 s

HOMEWORK KINEMATICS LESSON 7 PART 2
1. A long jumper leaves the ground with an initial velocity of 12 m/s at an angle of 28-degrees above the horizontal. Determine the time of flight, the horizontal distance, and the peak height of the long-jumper.
First, determine the coordinate system: The x-y coordinate at the take off point is (0,0).
Second, use trigonometric function to determine the initial velocity components:
Horiz: v ix
=(12 m/s) • cos (28 deg) =10.6 m/s
Vert: v iy
=(12 m/s) • sin (28 deg) = 5.6 m/s
Third, set up an x-y table, listing the known information.
Horizontal Info: v ix d xf
= ??? d ix
= 0
= 10.6 m/s
Vertical Info: d iy
= 0.0 m (total); d
y-peak
= ??? v iy
= 5.6 m/s v y-peak
= 0 m/s v fy
= -5.6 m/s a y
= -9.8 m/s/s a x
= 0 m/s/s
Fourth, use v fy
= v iy
+ a y
• t total
and vertical info to solve for time; the time of flight is 1.1 seconds (rounded from 1.1497 s).
Now use Δd x
= v ix
• t to solve for d xf
. Note that a x
is 0 m/s/s so the last term on the right side of the equation cancels. By substituting 10.6 m/s for v ix
and
1.1 s for t, the d xf
can be found to be 12 m (rounded from 12.1817 m)
Finally, use t up
= 0.5748 s and the equation d fy @ peak
- 0 = v iy
• t up
+ 0.5 • a y
• t up
2 with vertical info to find the y at the peak. Substituting and solving yields 1.6 m (rounded from 1.6193 m)

2. A football is kicked at an angle of 37.0 degrees above the horizontal with a velocity of 20.0 m/s. Calculate (a) the maximum height, (b) the time of travel before the football hits the ground, (c) how far away it hits the ground, (d) the velocity vector at the maximum height, and (e) the acceleration vector at maximum height . Assume the ball leaves the foot at ground level.
First, the x-y coordinate at the kick off point is (0,0).
Second, use trigonometric function to determine the initial velocity components:
Horiz: v ix
=(20.0 m/s) • cos (37 deg) =16.0 m/s
Vert: v iy
=(20.0 m/s) • sin (37 deg) = 12.0 m/s
Third, set up an x-y table, listing the known information.
Horizontal Info: v ix d xf
= ??? d ix
= 0
= 16.0 m/s a x
= 0 m/s/s
Vertical Info: d iy
= 0.0 m (total); d
y-peak
= ??? v iy
= 12.0 m/s v y-peak
= 0 m/s v fy
= -12.0 m/s a y
= -9.8 m/s/s
Fourth, Use equations to solve for unknowns:
(a) At max height, the velocity is horizontal, so v y
= 0, and this occurs at time: v fy
= v iy
+ a y
• t
0 = 12.0 + (-9.80)(t) t = 1.2244 s to the peak height

Using this time, we can determine the vertical distance the football travels to its peak height: d fy @ peak
- 0 = v iy
• t up
+ 0.5 • a y
• t up
2 d fy @ peak
= (12.0)(1.2244) + ½ (-9.80)(1.2244) 2 = 7.347 m to the peak of its height
(b) To find the time it takes for the ball to return to the ground, we can use the equation with the knowledge that v fy has the same magnitude but opposite sign to v iy v fy
= v iy
+ a y t
-12.0 = +12.0 +(-9.80)(t) t = 2.4489 s  2.45 seconds is the balls hang time.
(c) The total distance traveled in the x direction is found by applying the equation
Δd x
= v ix
• t
Δd x
= (16.0) (2.4489) v ix
= 39.1824

39.2 m/s
(d) At the highest point, there is no vertical component to the velocity.
There is only the horizontal component (which remains throughout the flight), so v x
= v peak
= 16.0 m/s
(e) The acceleration vector is the same at the highest point as it is throughout the flight, which is 9.80 m/s 2 downward.
3. A child sits upright in a wagon which is moving to the right at constant speed. The child ext3ends her hand and throws an apple straight upward
(from her point of view), while the wagon continues to travel forward at constant speed. If air resistnace is neglected, will the apple land (a) behind the wagon, (b) in the wagon, or (c) in front of the wagon? Explain.
Answer:

The child throws the apple straight up from her point of view with in initial velocity, v iy
. But when viewed by someone on the ground, the apple also has an initial horizontal component of velocity equal to the speed of the wagon, v ix
. Thus, to a person on the ground, the apple will follow the path of a projectile. Since the apple experiences no horizontal acceleration, v ix will stay constant and equal to the speed of the wagon. As the apple follows its arc, the wagon will be directly under the apple at all times because they have the same horizontal velocity. When the apple comes down, it will drop right into the wagon and into the outstretched hand of the child. The answer is (b).
4. A projectile is fired with an initial speed of 75.2 m/s at an angle of 34.5
o above the horizontal on a long flat firing range. Determine (a) the maximum height reached by the projectile, (b) the total time in the air, (c) the total horizontal distance covered, and (d) the velocity of the projectile
1.50 s after firing and its angle above the horizontal. Hint, find the vertical and horizontal velocities after 1.50 s and then use Pythagoras.
First, the x-y coordinate at the point of firing is (0,0).
Second, use trigonometric function to determine the initial velocity components:
Horiz: v ix
=(75.2 m/s) • cos (34.5 deg) =62.196 m/s
Vert: v iy
=(75.2 m/s) • sin (34.5deg) = 42.593 m/s
Third, set up an x-y table, listing the known information.
Horizontal Info: d xf
= ??? d ix
= 0 v ix a
= 62.196 m/s x
= 0 m/s/s
Vertical Info: d iy
= 0.0 m ;d
y-peak
= ??? v iy
= 42.593 m/s v y-peak
= 0 m/s v fy
= -42.593 m/s a y
= -9.8 m/s/s

(a) At the highest point, the vertical velocity, v ypeak
= 0. We find the maximum height from: v fy
2 = v iy
2 + 2a y
Δd, where Δd = d f
- d i
0 = 42.593
2 + 2(-9.8)(d fy
– 0) d yf
= 92.5593  92 .6 m
(b) Because the projectile returns to the same elevation, v fy
= -42.593 v fy
= v iy
+ a y t
-42.593 = 42.593 + (-9.8)t t = 8.6924

8.69 s
(c) We find the horizontal distance from
Δd x
= v ix
• t d fx
– 0 = 62.196*8.6924 d fx
= 540.63  5.40 x 10 2 m
(d) The horizontal velocity will be constant, v x
= 62.196 m/s. We find the vertical velocity from v fy
= v iy
+ a y t v fy
= 42.593 + (-9.8)(1.5) = 27.893 m/s
Therefore the magnitude of v i
is determined using pythagorus v i
2 = v ix
2 + v iy
2 v i
= 68.164  68.2 m/s
We find the angle from the triangle:
Tanθ = v yi
/v xi
= 22.254

22.2
o above the horizontal

5. The pilot of an airplane traveling 160 km/h wants to drop supplies to flood victims isolated on a patch of land 160 m below. The supplies should be dropped how many seconds before the plane is directly overhead?
Answer:
First choose a coordinate system with the origin at the landing point, (0,0) which makes v iy
= 640 m, and v ix
= ???
Second, set up an x-y table, listing the known information.
Convert 160 km/h to m/s

160 km/h *1000m/km* 1hr/3600 s =
44.444
Horizontal Info: v ix d xf
= 0 d ix
= ????
= 160 km/h
Vertical Info: d iy
= 640 m ; d
fy
= 0.00 v iy
= 0 m/s v fy
= ???? a y
= -9.8 m/s/s a x
= 0 m/s/s
Third, determine a strategy to solve problem. Because the horizontal velocity of the package is constant at the horizontal velocity of the airplane, the package is always directly under the airplane. The time interval is the time required for the fall, which we find from the vertical motion:
Δd y
= v iy t + ½ at 2 , where Δd = d fy
- d iy
0-640 = 0+1/2 (-9.8)(t 2 ) t = 5.7142  5.7 s