Chapter 3
Kinematics In 2
dimensions
Projectile motion
Objectives
At the end of this chapter the student will be able to:
Identify a projectile motion
Write the time equations of a projectile motion on
the y-axis and on the x-axis
Use the time equations to solve problems related to
projectile motion
What is A projectile motion?
Watch the following
video
Objective: Identify a projectile
Study of the motion of a projectile
At t = 0s, a projectile is launched with an initial velocity V0
making an angle θ with the + x-axis.
We consider that the origin coincides with the launching
point
Objective: Identify a projectile
Study of the motion of a projectile
Time equations
Along x-axis:
ax = 0
Vx = V0. Cos θ
x = V0 . t . Cos θ
Objective: Write the time equations of a projectile motion on the y-axis
Along y-axis:
ay = -g
Vy = -gt + V0 . sin θ
y = - ½ .g.t2 + V0 . t
. sin θ
Study of the motion of a projectile
Along x-axis:
ax = 0
Vx = V0. Cos θ
x = V0 . t . Cos θ
Along y-axis:
ay = -g
Vy = -gt + V0 . sin θ
y = - ½ .g.t2 + V0 . t . sin θ
Objective: Write the time equations of a projectile motion on the y-axis
Study of the motion of a projectile
• To find the maximum height
(hmax):
y=0
hmax
We use Vy = 0, we find t then we
insert t in y.
• To find the range:
We use y = 0, we find t
then we insert t in x.
Range
Along y-axis:
ay = -g
Vy = -gt + V0 . sin θ
y = - ½ .g.t2 + V0 . t .
sin θ
Objective: Write the time equations of a projectile motion on the y-axis
Along x-axis:
ax = 0
Vx = V0. Cos θ
x = V0 . t . Cos
θ
Study of the motion of a projectile
• To find the Speed of a projectile
at a certain instant we fin Vx then
Vy then we find V:
Along y-axis:
ay = -g
• To find the direction
of the velocity:
Objective: Write the time equations of a projectile motion on the y-axis
Vy = -gt + V0 . sin θ
y = - ½ .g.t2 + V0 . t .
sin θ
Along x-axis:
ax = 0
Vx = V0. Cos θ
x = V0 . t . Cos
θ
Application 1
Use the given on the figure to find:
1- The time equations of the projectile
2- The coordinates of the highest point.
3- The range.
4- The speed when the projectile reaches the ground.
Objective: Use the time equations to solve problems related to
Application 1
1- Time equations:
Vx = V0. cosθ = 9.6
Vy = -gt + V0 Sinθ = -10t +
11.5
X = V0.t.cosθ = 9.6t
Y= - ½ gt2 + V0 . t . Sinθ = -5t2
+ 11.5t
Objective: Use the time equations to solve problems related to
2- Highest point: Vy = 0
-10t + 11.5 = 0 so t = 1.15 seconds
So the projectile took 1.15 seconds to reach the
highest point.
X = 9.6t = 9.6 x 1.15 = 11 m
Y= -5t2 + 11.5t = -5 (1.15)2 +11.5x1.15 = 6.6
m
So the coordinates of the highest point are:
X = 11 m and Y = 6.6 m
Application 1
3- To find the range we use Y =
0.
-5t2 + 11.5t = 0
t(-5t+11.5) = 0
4- To find the speed we find:
t = 11.5/5 = 2.3 seconds
Now we replace t in X.
Vx = 9.6 m/s
X = 9.6 x 2.3 = 22 m
Vy = -10 x t + 11.5 = -10x2.3
+11.5 = -11.5 m/s
= 15m/s
Objective: Use the time equations to solve problems related to
Application 1
Now click on the image an verify you answers
Objective: Use the time equations to solve problems related to
Application 2
Use the given on the figure to find:
1- The time equations of the projectile
2- The time the projectile takes to reach the ground .
3- The horizontal distance between the canon and the impact
Objective: Use point.
the time equations to solve problems related to
Application 2
1- Time equations:
Vx = V0. cosθ = 15
Vy = -gt + V0 Sinθ =
-10t
X = V0.t.cosθ = 15t
Y= - ½ gt2 + V0 . t .
Sinθ = -5t2
Objective: Use the time equations to solve problems related to
θ = 0˚
A
2- Let’s name the impact
point A
YA = -10 m
2
So,
-10
=
-5t
and
t
=
3- To find the horizontal
1.4seconds
distance we need to find XA.
X = 15 x 1.4 = 21 m
Application 2
Now click on the image an verify you answers
Objective: Use the time equations to solve problems related to
Remark
Watch the following video
Objective: Use the time equations to solve problems related to
Remark
Show using the time equations that the 2 objects
will reach the ground at the same time
Objective: Use the time equations to solve problems related to
End Of Chapter 3
Thank you for your
attention