Younes Sina's Lecture at Pellissippi State Community College

1. Chapter 3
Motion in Two Dimensions
Younes Sina

2. Curved motion of a projectile is a motion in two
dimensions. For example, if an object is to move along a
curve as shown in the following figure, its position and
its velocity at any given time ( t ) can be determined by
its two coordinates ( x ) and ( y ).
This parabolic motion
can be replaced by a
constant speed motion
in the horizontal
direction (x-axis) and
an accelerated motion
in the vertical direction
(y-axis).

3. The package falling from the plane is an example of
projectile motion.
An accelerated motion in the vertical direction
constant speed motion in the horizontal direction (x-axis)

4. The path of motion of the ball in air (neglecting
air resistance) is a parabola.

5. The reason for the constant speed motion along the x-
axis is that there is no gravitational pull horizontally to
accelerate the ball and, neglecting air resistance, there is
no force to slow it down. In the y-direction, because of
the gravitational pull, it first slows down as it goes
upward, and then speeds up as it returns to the ground.

6. In the x-direction: In the y-direction:
x = (1/2) ax t2 + vix t y = (1/2) g t2 + viy t
Since ax = 0, g = (vfy -viy )/ t
x = vix t vfy
2 - viy
2 = 2 g y

8. Example : A cannon ball is fired at an initial speed of 250m/s
through a 27o angle. Determine
(a) its horizontal and vertical components of velocity,
(b) the time it takes to reach the highest point,
(c) the highest elevation it reaches,
(d) the time it spends in air before landing,
(e) the farthest horizontal distance it travels before landing
(f) the equation of its path.

9. In the x-direction: In the y-direction:
x = (1/2) ax t2 + vix t y = (1/2) g t2 + viy t
Since ax = 0, g = (vfy -viy )/ t
x = vix t vfy
2 - viy
2 = 2 g y
Vix = v0cos(θo)
Viy=v0cos(θo)
Vfy = 0
x = vix t

10. Solution:
(a) Vix = 250cos(27o) = 223m/s
Viy = 250sin(27o) = 113m/s.
(b) At the highest point, Vfy = 0 ; from g = (Vfy - Viy) / t
t = (Vfy - Viy) /g = (0 - 113 ) / -9.8 = 12s
(c) At the highest point, Vfy = 0 ; from (Vfy)2 - (Viy)2 = 2 g y
02 - 1132 = 2(-9.8)y ; y = 650m
(d) Using the result of Part (b), ttotal = 2(12s) = 24s.
(e) Using the only x-equation, x = Vix t,
x = ( 223 m/s)( 24s ) = 5400m
(f) To find the equation of the path (a parabola), the common variable or the
common parameter ( t ), must be eliminated between the x- and y-components of
motion. To do this, let us first write down the equations of motion in the x- and y-
directions.
In the y-direction, y = (1/2) g t2 + Viyt
y = -4.9t2 + 113t
In the x-direction, x = Vix t
x = 223 t or t = x / 223
Substituting for ( t ) in the y-equation, yields:
y = -4.9( x / 223)2 + 113 ( x / 223)
y = -0.000099 x2 + 0.51 x (A parabola)

11. Example : A small ball is rolling at a speed of 5.0m/s on a horizontal
table 1.2m high with respect to the floor.
How far from the edge of the table does it land after rolling off the table?
V= 5.0m/s
h=1.2m
θ= 0o

13. Homework
problems 1 through 3 chapter 3