2D Motion 2015 cjcb Angled Projectile Motion

RECAP: What is Projectile Motion? Projectile motion is the motion of an object that is being thrown or launched. The object may be launched horizontally or at an angle. IN THE ABSENCE OF FRICTION OR AIR RESISTANCE, the ONLY FORCE ACTING ON A PROJECTILE IS THE FORCE OF GRAVITY !!!! ***** (whether thrown straight up, horizontally, or at an angle)

What Actually Happens?

What Does This Look Like? KEY POINTS: (1)HORIZONTAL and VERTICAL motion are treated separately (2)HORIZONTAL motion is a constant velocity problem (3)VERTICAL motion is like a free fall problem (acceleration, with a = g)

Solving Problems: HORIZONTAL PROJECTILE MOTION Horizontal: (constant velocity) v x =d x /t Vertical: (free fall) d y = v yi t + 1/2 gt 2 v yf 2 = v yi 2 + 2gd y v yf = v yi + gt Isolate the horizontal and vertical components into separate sides and solve for each piece.

EXAMPLE: (Worksheet Prob #1 w/ extra parts) Erica kicks a soccer ball at 12 m/s horizontally from the edge of a roof of a building which is 30.0 m high. QUESTIONS: (a) WHEN does it strike the ground? ( t = ? ) (b) How far from the building does it land? ( d x = ? ) (added question) (c) With what velocity does it strike the ground? ( v xf & v yf )

Picture: HORIZ:VERT: knowns: v xi = 12 m/s d y = 30.0 m (or d V ) and also: v yi = 0 m/s v xf = 12 m/s a = g = 9.8 m/s 2 unknowns: d x = ? (or d H ) v yf = ? t = ? (time is same on both sides) t = ? STEP 1: DRAW PICTURE (down and out is POSITIVE) STEP 2: IDENTIFY KNOWNS & UNKNOWNS

STEP 3: WRITE EQUATIONS – based on knowns & unknowns & identify where have enough information to solve for a variable HORIZ:VERT: knowns: v i = v x = 12 m/s d y = 30.0 m v yi = 0 m/s a = g = 9.8 m/s 2EQUATIONS: v x =d x /t d y = v yi t + 1/2 gt 2 2 unknowns: t & d x only 1 unknown: t so can’t solve YET use this to SOLVE for time (also know enough to solve for v yf in part c): v yf 2 = v yi 2 + 2gd y

Erica kicks a soccer ball at 12 m/s horizontally from the edge of a roof of a building which is 30.0 m high. QUESTIONS: (a) WHEN does it strike the ground? ( t = ? ) STEP 4: START SOLVING where have enough info so to do 1 st = TIME on VERTICAL SIDE HORIZ:VERT: knowns: v i = 12 m/s d y = 30.0 m v yi = 0 m/s a = g = 9.8 m/s 2EQUATIONS: v x =d x /t d y = v yi t + 1/2 gt 2 (wait to work on this side) since v yi =0 simplify and rewrite as: t = (2 d y / g) 1/2 t = (2 * 30.0 m) / 9.8 m/s 2 ) 1/2 t = 2.47 s

Erica kicks a soccer ball at 12 m/s horizontally from the edge of a roof of a building which is 30.0 m high. QUESTIONS: (b) How far from the building does it land? ( d x = ? ) Now NOTICE that TIME is in PURPLE. That’s because time is the SAME in the HORIZ. and VERT. directions. SO, t calculated using equations on the vertical side can be used in equations on the horizontal side and vice versa. HORIZ: VERT: knowns: we solved for: v i = 12 m/s t = 2.47 s * Because constant velocity v i = v x = v x = 12 m/s EQUATIONS v x =d x /t - rearrange: d x = v x * t d x = 12 m/s * 2.47 s d x = 29.7 m from building

Erica kicks a soccer ball at 12 m/s horizontally from the edge of a roof of a building which is 30.0 m high. QUESTIONS: (c) With what velocity does it strike the ground? ( v xf & v yf ) Now for part “c”. We KNOW the horizontal velocity when it lands because it’s the same as the start (constant velocity). Use kinematic equations to solve for v yf. HORIZ: VERT: Back to knowns & unknowns: v xi = v xf = v x = 12 m/s d y = 30.0 m v yi = 0 m/s a = g = 9.8 m/s 2 t = 2.47 s *solved for in part (a) EQUATIONS: v yf 2 = v yi 2 + 2gd y ( independent of t) OR v yf = v yi + gt ( using t from “part a” ) one way: v yf = 0 m/s + ( 9.8 m/s 2 * 2.47 s) v yf = 24.3 m/s

Solving Problems: ANGLED PROJECTILE MOTION NOW: What if you shot the cannon UP (or down) at an angle instead of horizontally? It would look something like this:

Solving Problems: ANGLED PROJECTILE MOTION Angled Projectile Motion can START at the ground OR “in the air” (at a different height than the final height) But The Trajectory Is Always Parabolic!!  symmetry of motion up and down

For ANGLED PROJECTILE MOTION: Still have constant v x, BUT it is no longer equal to the overall initial velocity Still have a = g BUT no longer v yi = 0 m/s - only acceleration is “g”, in vertical direction There is an initial velocity in BOTH the HORIZONTAL and VERTICAL directions - but can calculate these two components - just like last unit, use vector math (cos and sin) to resolve velocity into v x and v yi

Solving Problems: ANGLED PROJECTILE MOTION Horizontal: (constant velocity) v x = v i  cos  Vertical: (constant acceleration) v yi = v i  sin  OUR NEW “STEP 1” is to break up the initial velocity into its “x” (horizontal) and “y” (vertical) components: V i = 20 m/s THEN Solve as for HORIZONTAL PROJECTILE MOTION, using the v x and v yi calculated here.

Solving Problems: ANGLED PROJECTILE MOTION Horizontal: (constant velocity) v x = v i  cos  v x = 50  cos60° v x = 25 m/s Vertical: (constant acceleration) v yi = v i  sin  v yi = 50  sin60° v yi = 43 m/s EXAMPLE: Break up the initial velocity into its “x” (horizontal) and “y” (vertical) components where v x = 50 m/s and  = 60° :

Solving Problems: ANGLED PROJECTILE MOTION Horizontal: (constant velocity) v x =d x /t Vertical: d y = v yi t + 1/2 gt 2 v yf 2 = v yi 2 + 2gd y v yf = v yi + gt For the rest of the problem, you STILL Isolate the horizontal and vertical components into separate sides and solve for each piece. SAME EQUATIONS as for HORIZONTAL

Solving Problems: ANGLED PROJECTILE MOTION A BIT ABOUT “VOCABULARY”: - “ HANG TIME ” refers to the time in the air (t) -“ RANGE ” refers to the horizontal distance travelled (d x ) - “ MAXIMUM HEIGHT ” is the height where the vertical velocity is zero ( v y = 0 m/s )

Solving Problems: ANGLED PROJECTILE MOTION EXAMPLE: p. 152 #6 A rock is thrown from a 50.0 m high cliff with an initial velocity of 7.0 m/s at an angle of 53.0 ° above the horizontal. Find the velocity vector for when it hits the ground. How long does it take?

STEP 1: DRAW PICTURE (UP and out is POSITIVE)\ STEP 2: Calculate the initial Velocity in the x (horizontal) and y (vertical) directions, given v i = 7.0 m/s and angle of the projectile,  = 53.0° (alternatively can substitute expressions for v x and v yi into basic equations) Picture: Calculating initial velocity: : Vertical: (constant acceleration) v yi = v i  sin  v yi = 7.0  sin53.0° v yi = 5.59 m/s Horizontal: (constant velocity) v x = v i  cos  v x = 7.0  cos53.0° v x = 4.21m/s

STEP 3: IDENTIFY KNOWNS & UNKNOWNS HORIZ: VERT: knowns: knowns: v x = v i  cos  = 4.21 m/s  v yi = v i  sin  = 5.59 m/s d y = m a = g = m/s 2 unknowns: d x = ? (or d H ) v yf = ? t = ? (time is same on both sides) t = ? ** DOES THIS LOOK FAMILIAR? [See the HORIZONTAL PROJECTILE MOTION example]

STEP 4: WRITE EQUATIONS – based on knowns & unknowns & identify where have enough information to solve for a variable HORIZ:VERT: knowns: v x d y, v yi, and a = gEQUATIONS: v x =d x /t v yf 2 = v yi 2 + 2gd y only 1 unknown: v yf 2 unknowns: t & d x d y = v yi t + 1/2 gt 2 - this problem doesn’t ask for only 1 unknown: t t & d x, but you could solve for - could use this to SOLVE for time them as in the earlier example of horizontal motion (t from VERT then substitute t into equation above)

STEP 4a: SOLVE FIRST: FIND EACH INDIVIDUAL COMPONENT OF FINAL VELOCITY: HORIZ: VERT: v xi = v xf = v x = 4.21 m/s v yf 2 = v yi 2 + 2gd y ( independent of t) Substituting: v yf 2 = [v i  sin  2 + 2gd y = 5.59 m/s v yf 2 = [5.59 m/s  2 + 2(-9.8 m/s 2 )(-50.0 m) v yf = 31.8 m/s

STEP 4b: SOLVE SECOND: COMBINE INTO SINGLE VELOCITY VECTOR – ADDING VECTORS AS IN THE LAST UNIT: use Pythagorean Theorem SO: V f 2 = v x 2 + v yf 2 V f 2 = ( 4.21 m/s) 2 + ( 31.8 m/s ) 2 V f = m/s AND tan  = v yf / v x tan  = 31.8 m/s/ 4.21 m/s  = tan -1 (31.8 / 4.21 )  = 82.45° ** We will also talk about solving for RANGE, HANG TIME, and MAXIMUM HEIGHT **

Homework Angled Projectile Motion Worksheet, #1-3 Page 152, #4-6