Projectile Motion 2
Launch Angles
(symmetrical and
asymmetrical trajectories)
Physics 12
Comprehension Check

An arrow is launched from a cliff with an
initial velocity of 25m/s. If the cliff is 25m
high determine:
 The
time in the air
 How far from the base of the cliff it hits
 How fast it is going when it hits the ground
Projectile Diagram – Launched at
an Angle
Projectiles Launched at an Angle
Both vi and vf have x and y components
 Still follow a parabolic trajectory

Symmetrical Trajectories




When you launch a projectile from the same
vertical height that it lands
dy = 0
Think back to physics 11… what do you know
about the speed/velocity of an object that goes
the same distance up as it goes down?
viy = -vfy
Symmetrical Trajectories

What about the overall time
and the time to its peak?
 Time
to peak/vertex = ½ time
in air

What do you think about the
angle it is launched at and
the angle it lands at?
 θi
= θf
What about the x-direction?

Is there any force acting on the projectile
once it is launched? Acceleration?
 No
force acting on the projectile once
launched except gravity, which only affects
the y-direction.

What can we say about vix and vfx?
 If
ax = 0, then vix and vfx
Hint…

Remember: To find the initial velocity in the x
and y direction, you must use trigonometry!

X direction: cosine
Y direction: sine


Remember that at its peak, an object’s y velocity
is 0 because it is changing from a positive
velocity to a negative velocity (grade 11).
Example 1

A player kicks a football with an initial
velocity of 29 m/s at 69.0°. Assuming it
lands at the same height from which it was
kicked, determine the time of flight.
Projectiles Launched at an Angle

A golfer uses a club that
launches a golf ball at a 15°
angle at a speed of 45m/s.
Determine the following:
 The
time the golf ball is in the air
 The horizontal distance the ball
travels
 The velocity as it strikes the ground
 The maximum height the ball
attains



Start with the
position in the
y equation
Include the
initial velocity
in the y term
Solve for
when the
position in the
y is equal to
zero
2



gt
d y (t ) 
 voyt  d oy
2
2 2
0  4.91m / s t  45m / s sin( 15)t  0
4.91m / s t  45m / s sin( 15)t
2 2
4.91m / s t  45m / s sin( 15)
2
t  2 .4 s


Use the time
from the
previous
question and
the position in
the x equation
Solve for the
range
(horizontal
distance)



d x (t )  voxt  d ox

d x (2.4 s )  45m / s cos(15)( 2.4 s )  0

2
d x (2.4 s )  1.0 x10 m

Solve for the velocity in the x and y direction


v x (t )  vox

v x  45m / s cos(15)

v x  44m / s

 
v y (t )  gt  voy

v y (2.4 s )  9.81m / s 2 (2.4 s )  45m / s sin( 15)

v y (2.4 s )  12m / s

Use Pythagorean Theorem and Trig to solve for
final velocity

v x  44m / s

v y (2.4 s )  12m / s
v  (44m / s )  (12m / s )
2
v  45m / s
1 12 m / s
  tan
44m / s
  15o

o
v  45m / s,15
2

Max height will occur when y velocity is equal to
zero. Solve for time and then sub into y position
equation

 
v y (t )  gt  voy
0  9.81m / s 2t  45m / s sin( 15)
t  1.2s
2



gt
d y (t ) 
 voyt  d oy
2

d y (1.2s)  4.91m / s 2 (1.2s) 2  45m / s sin( 15)(1.2s)

d y (1.2s)  6.9m
Comprehension Check

A cannonball is launched with an initial
velocity of 25m/s at an angle of 32°.
 How
long is it in the air?
 What is the horizontal distance that the
cannonball travels?
 What is the maximum height?
 With what velocity does it strike the ground?
 At 56m, there is a 2.0m high wall; does the
cannonball clear the wall?
Comprehension Check
2.7s
 57m
 8.9m
 25m/s, 32°
 No

Practice Problems

Page 536
 Questions

1-8
Page 549
 Questions
13-14
Asymmetrical Trajectories

When the launch height is different from the
landing height
 Launching
a projectile at an angle from the top of a
hill
 Launching a projectile from ground level into water




Final and initial y velocities will not be the same
Final and initial x velocities will be the same
Time to the vertex/peak will not be ½ time
Vertical velocity will still be 0 at peak
Things to remember…


You will need to use the quadratic equation
and…
If the projectile lands HIGHER than the launch
site:
 There
will be 2 positive roots but the first root will be
the time it takes to get to that height on the way up
(so use the other value)

If the projectile lands LOWER than the launch
site:
 There
will be a 1 positive root (the negative root will
be before the launch happened)
Example 1

A golfer hits the golf ball off the tee, giving it an
initial velocity of 32.6 m/s at an angle of 65’ with
the horizontal. The green where the golf ball
lands is 6.30m higher than the tee. Neglect air
friction. Find:
 The
time the golf ball is in the air
 How far it travels horizontally
 The velocity of the ball just before it hits the ground
Try this…





A golfer strikes a ball with a velocity of 470 m/s
at 35.0’. If the ball lands on the fairway 3.50m
below the tee, find the
Time of flight (5.63s)
Range (217 m)
Max height (37.0 m)
Velocity at end of flight (47.7 m/s [E 36.2’S])
Try this…

A golfer stands on an elevated tee that is
6.50m higher than the green. If the ball
next to the cup (183m away) and the ball
is hit at a 40.0° angle, determine:
 The
speed at which the ball left the club (41.8
m/s)
 The time the golf ball is in the air (5.71s)
 The maximum height the ball attains (36.8m)