Two-Dimensional Motion Chapter 3

A little vocab  Projectile = any object that moves through space acted on only by gravity  Trajectory = the path followed by a projectile  Range = the horizontal distance a projectile travels

Puzzle  If a ball is dropped straight down at the same time one is launched horizontally, which ball will hit the ground first?  Does this work the same if it is a bullet launched horizontally, at the same time a ball is dropped from the same height?

Without any means to propel an object, once it is moving horizontally, it has constant horizontal motion… x = vt (no acceleration) In free fall, vertical motion… y = v i t + ½ gt 2 We know…

Put these motions together

Motion in Two Dimensions  Using + or – signs is not always sufficient to fully describe motion in more than one dimension Vectors can be used to more fully describe motion  Still interested in displacement, velocity, and acceleration

Projectile Motion  An object may move in both the x and y directions simultaneously It moves in “two dimensions”  The form of two dimensional motion we will deal with is an important special case called projectile motion

Assumptions of Projectile Motion  Because this is a “perfect physics world” We may ignore air resistance and friction We may ignore the rotation of the earth  With these assumptions, an object in projectile motion will follow a parabolic path

Rules of Projectile Motion  The x- and y-directions of motion are completely independent of each other  X-motion and Y-motion happen at the same time  a y = g Why? The y-direction is free fall  a x = 0 Why? The x-direction is uniform motion There is nothing else pushing in the horizontal direction

Horizontal Projectile Motion  something thrown horizontally Without gravity, it would continue along a horizontal path. Because gravity acts “downward”, the path is half of a parabola. Start at maximum height Given a push in the X-direction only

Solving Horizontal Projectile Motion  Make a chart with 2 sides  What do we ALREADY know (without even having a problem to solve?) HorizontalVertical a = v i = v f = t = x = a = v i = v f = t = x = 0 # # = v i # # = range 9.8m/s 2 = g 0 # (just before it hits) # Max height = y NOTE: Can only cross over at t!

Solving contunued…  Use kinematics equations to solve  Horizontal… if a = 0, the only equation that matters is #2 x = v i t  Vertical … 1.v f = v i + gt 2.y = v i t + ½gt 2 3.v f 2 = v i 2 + 2gy HINT: Solve for time first… goes on both sides of chart!

Example A dart player throws a dart horizontally at a speed of 12.4 m/s. The dart hits the board 32 cm below the height from which it was thrown. How far away is the player from the board? HorizontalVertical a = v i = v f = t = x = a = v i = v f = t = y = m/s ? ? 9.8m/s 2 0 ? ? 32cm =.32m

What can we solve first?  We are looking for “x”  Not enough info on the horizontal side…  On the vertical side, we can solve for v f t  Solve for t, so we can use it on the horizontal side  Don’t really care about v f this time

The math…  For t y = v i t + ½gt 2.32 = 0 + ½ (9.8)t 2 t = s HorizontalVertical a = v i = v f = t = x = a = v i = v f = t = y = m/s s ? 9.8m/s 2 0 ? s 32cm =.32m  For x x = vt x = (12.4)(.2556) x = 3.17m

What if…  the previous question asked for the total final velocity?  This is where VECTORS come in…  We knew that v f was the same as the v i in the horizontal = 12.4m/s  We can find v f in the vertical v f = v i + gt v f = 0 + (9.8)(0.2556) v f = 2.5m/s

We aren’t done yet.  Total velocity is a vector in polar coordinates. 12.4m/s 2.5m/s Resultant (r,Θ) r 2 = r = 12.6m/s tanΘ = -2.5/12.4 Θ = tan -1 ( ) Θ = o Θ