Chapter 3

Motion in Two Dimensions

Younes Sina

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).

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)

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

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.

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

2 - viy2 = 2 g y

Example : A cannon ball is fired at an initial speed of 250m/sthrough 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.

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

2 - viy2 = 2 g y

Vix = v0cos(θo) Viy

= v 0

cos(θ

o)

Vfy = 0

x = vix t

Solution:(a) Vix = 250cos(27o) = 223m/sViy = 250sin(27o) = 113m/s.(b) At the highest point, Vfy = 0 ; from g = (Vfy - Viy) / tt = (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 + Viyty = -4.9t2 + 113tIn the x-direction, x = Vix tx = 223 t or t = x / 223Substituting 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)

Example : A small ball is rolling at a speed of 5.0m/s on a horizontaltable 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

Homeworkproblems 1 through 3 chapter 3