4 How are they different? Projectiles Launched Horizontally The initial vertical velocity is 0.The initial horizontal velocity is the total initial velocity.Projectiles Launched At An AngleResolve the initial velocity into x and y components.The initial vertical velocity is the y component.The initial horizontal velocity is the x component.
5 Some Details About the Rules x-directionax = 0vx = vix = vicosΘi = constantx = vixtThis is the only equation in the x-direction since there is constant velocity in that directionInitial velocity still equals final velocity
6 More Details About the Rules y-directionviy = visinΘiFree fall problema = gObject slows as it goes up (-9.8m/s2)Uniformly accelerated motion, so the motion equations all holdJUST LIKE STOMP ROCKETSSymmetrical
7 Problem-Solving Strategy Resolve the initial velocity into x- and y-componentsTreat the horizontal and vertical motions independentlyMake a chart again, showing horizontal and vertical motionChoose to investigate up or down.Follow rules of kinematics equations
8 Solving Launched Projectile Motion vi = (r , Θ) = (vix , viy)Horizontal Verticala =vi =vf =t =x =vix## = rangea =vi =vf =t =x =+/- 9.8m/s2 = g0 or viyviy or 0#Max height = yUP or DOWN INVESTIGATION… Where do the resolved components go?
10 Example 1:The punter for the Steelers punts the football with a velocity of 27 m/s at an angle of 30. Find the ball’s hang time, maximum height, and distance traveled (range) when it hits the ground. (Assume the ball is kicked from ground level.)Looking for:Total time (t)Max height (y)Range (x)Given:vi = (27m/s, 30o)
11 What do we do with the given info? vi = (27m/s, 30o)What are the units?vi = (23.4m/s, 13.5m/s)“resolved” vectorm/s27m/s30oViy = 27sin30Viy = 13.5Viy = 13.5m/sVix = 27cos30Vix = 23.4Vix = 23.4m/s
12 So where does this info “fit” in the chart? Horizontal Verticala =vi =vf =t =x =23.4m/sa =vi =vf =t =x =Viy if solving “up” = 13.5m/sViy if solving “down” = 13.5m/s
17 On the horizontal side Max height occurs midway through the flight. We found t = 1.38s both directions (up and down).How long is the projectile in the air? DOUBLE this time for total air time t = 1.38 x 2 = 2.76sWhat about range? x = vt x = (23.4)(2.76) x = 64.6m = RANGE
18 This tells us…Now we only need an initial velocity vector to determine all of the information we need to have a detailed description of where an object is in its path.vi = (r , Θ) = (vix , viy)
19 Maximum Range vs. Maximum Height What angle of a launched projectile gets the maximum height?What angle of a launched projectile gets the maximum range?90o45o
20 Projectile Motion at Various Initial Angles Complementary values of the initial angle result in the same rangeThe heights will be differentThe maximum range occurs at a projection angle of 45o
21 Non-Symmetrical Projectile Motion Follow the general rules for projectile motionBreak the y-direction into partsup and downsymmetrical back to initial height and then the rest of the height
22 your homework …We are going to see what kind of job Hollywood writers and producers would do on their NECAP assessments…Watching a clip of the Bus Jump, use the timer provided to time the flight of the bus and then do the actual calculations for homework.