Projectile Motion (Summary) Projectile Motion Vertical Velocity ...

Projectile Motion (Summary)• A projectile is an object in free fall: subject togravity and air resistance• In this context, we will ignore the effects of airresistance; sometimes, this isnot a good idea• When air resistance is ignored, projectiles area good example of constant acceleration• Projectile position, displacement, velocity, andacceleration can be better understood usingthe equations of constant accelerationGalileo’s equations of constant acceleration:v 2 = v 1 + atd = v 1 t + ½at 2v 22 = v 12 + 2add = displacementv = velocitya = accelerationt = timesubscript 1 = initial timesubscript 2 = final timeYou may also use the equations below (derivations ofthe aforementioned equations). In many instances,they can save you time, but are not necessary.2(V1)Maximal Height =− 2g(2V1)Time of Flight =- g2- V1- V1- 2ghTime of Flight =gProjectile MotionWith no air resistance, the path followed by aprojectile will be parabolicVertical VelocityHorizontal and vertical components of projectilemotion should be treated separatelyGravity will cause the vertical component ofvelocity to change during the flightverticalvelocity = 0Horizontal VelocityHowever, horizontal velocity is constantthroughout flight. Why?Projectile MotionIn review, if air resistance is negligible:• Horizontal velocity is constant• Vertical velocity changes by -9.81 m/s/sV VVhttp://www.youtube.com/watch?v=R6pPwP3s7s4&feature=player_embeddedhttp://www.youtube.com/watch?v=PvC33gSTSR4&feature=relatedV Hhttp://www.youtube.com/watch?v=37qyvTRVus81

Vertical Projectile MotionVertical motion of a projectile (e.g., a water balloonthat is dropped from the edge of a building) is verypredictabled = v 1 t + ½at 2v 2 = v 1 + at0 m-10 m-20 mVertical Projectile Motiond = v 1 t + ½at 2and v 2 = v 1 + at0 m-10 m-20 mtime position velocity acceleration0 s-30 m-40 mtime position velocity acceleration0 s 0 m 0 m/s −9.8 m/s 2-30 m-40 m1 s-50 m1 s-50 m2 s-60 m2 s-60 m3 s3 sVertical Projectile MotionVertical Projectile Motiond = v 1 t + ½at 2and v 2 = v 1 + at0 md = v 1 t + ½at 2and v 2 = v 1 + at0 m-10 m-10 m-20 m-20 mtime position velocity acceleration0 s 0 m 0 m/s −9.8 m/s 2-30 m-40 mtime position velocity acceleration0 s 0 m 0 m/s −9.8 m/s 2-30 m-40 m1 s −4.9 m −9.8 m/s −9.8 m/s 2-50 m1 s −4.9 m −9.8 m/s −9.8 m/s 2-50 m2 s-60 m2 s −19.6 m −19.6 m/s −9.8 m/s 2-60 m3 s3 sVertical Projectile Motiond = v 1 t + ½at 2and v 2 = v 1 + at0 mAnother exampleHow high was the cliff (what is d V )?-10 m-20 mtime position velocity acceleration0 s 0 m 0 m/s −9.8 m/s 21 s −4.9 m −9.8 m/s −9.8 m/s 22 s −19.6 m −19.6 m/s −9.8 m/s 23 s −44.1 m −29.4 m/s −9.8 m/s 2-30 m-40 m-50 m-60 m2

Another exampleHow high was the cliff (what was d V )?t DOWN = ~ 2 s, and usingd V = V 1 t + ½at 2 , we learnedthat:d V = 0 + ½ · -9.81m/s 2 · 2s 2 ,-19.6 mPerhaps, more importantly, what was V F ?V F = 0 + -9.81m/s 2 · 2s = 19.62 m/s, or ~44 mphHorizontal Projectile MotionThe horizontal displacement, or range of aprojectile, is the main performance index inmany cases.If air resistance is considered to be negligible,there is no net force in the horizontal direction(ΣF x = 0; a x = 0)Given the equation: d = v 1 t + ½at 2 ,we can assume that: d X = V X × tHorizontal VelocitySo, d X = V X × t, allows us to calculate horizontalrange for a projectile, if we know the horizontalvelocity and air time. First, what affectshorizontal velocity?Flight TimeNext, what affects flight time? (two primary factors)1. Vertical speed at release: affected byangle of release and applied force2. Height of release(if takeoff height = landing height, then t UP = t DOWN )t UPt DOWNSome ApplicationBased upon what we havelearned, what factors can onemanipulate to influence thehorizontal displacement, orrange, of a projectile? What ismost important?Relative Height of ReleaseSpeed of ReleaseAngle of ReleaseWhat is most important?Does increasing the height of release alwayslead to greater d x ?Yes, why?d x = v x × tDoes increasing speed of release always leadto greater d x ?Yes, but again, why?d x = v x × t3

What is most important?What about angle of release?When height of release is 0 m, optimal anglefor maximum d x is 45°As height of release ↑, optimal angle ↓d x = v x × tWhat is most important?d H is most sensitive to speed of release…d x = v x × tSpeed of release can positively affect v H and t TOTWhat are the performance-relatedimplications of this information?More Application…How do actual release angles compare totheoretical optima?Shot Put:• Positive height of release, so optimal angleshould be slightly lower than 45°• Theoretically optimal angle is about 40-41°More Application…Long Jump:Positive height of release, so optimal angleshould be slightly lower than 45°• Skilled shot-putters use angles of 31-36°• Close, but why the discrepancy?Projectile Motion:Theory vs RealityLong Jump:• Theoretically optimal angle is about 42°• Top long jumpers use angles of 17-23°• Very different. Why the major discrepancy?When traveling at ~10 m/s, there is not enoughtime to generate a large takeoff angleLong jumpers sacrifice optimal angle to getmaximum speed.Projectile Motion (Summary)• A projectile is an object in free fall: subject togravity and air resistance• In this context, we will ignore the effects of airresistance; sometimes, this isnot a good idea• When air resistance is ignored, projectiles area good example of constant acceleration• Projectile position, displacement, velocity, andacceleration can be better understood usingthe equations of constant acceleration4

Some Practice: The Shot PutHow far will the shot travel?Remember: d H = v H × tCase Ispeed of release = 12 m/sheight of release = 0 mangle of release = 30°0 m30°12 m/sSome Practice: The Shot PutUse d H = v H tneed to know v H and tv H = v cos θ = 12 m/s ⋅ cos 30° = 10.4 m/sWhat is total time in the air (t TOT )?If height of release = 0, then t UP = t DOWNand t TOT = t UP + t DOWNWhat else do we know?For upward part of flight:v 1 = v V = v · sin θ = 12 m/s ⋅ sin 30° = 6 m/sv 2 = 0 m/s a = −9.81 m/s 2Some Practice: The Shot PutFind appropriate equation for constant accelerationv 2 = v 1 + atPlug in v 1 , v 2 , and a, then solve for t:0 m/s = 6 m/s + (−9.81m/s 2 )(t UP )t UP = 0.61 sSome Practice: The Shot PutLets use a more realistic height of releaseCase IIspeed of release = 12 m/sheight of release = 2.1 mangle of release = 30°t TOT will be 2 × t UP : t TOT = 2 × 0.61 s = 1.22 s (Why ?)d H = v H t TOT = 10.4 m/s × 1.22 s = 12.7 mSo, the shot traveled 12.7 m horizontally2.1 m30°12 m/sSome Practice: The Shot PutWe still use d H = v H t , but now t UP ≠ t DOWNv H is found the same way as before:v H = v cos θ = 12 m/s ⋅ cos 30° = 10.4 m/sand t UP is also the same (using v 2 = v 1 + at):0 m/s = 6 m/s + (−9.81m/s 2 )(t UP )t UP = 0.61 sSo how do we calculate t DOWN ?We need to find the upward (d UP ) anddownward (d DOWN ) displacementsSome Practice: The Shot PutFind another appropriate equationd = v 1 t + ½at 2Calculate the upwards displacement:d UP = (6 m/s)(0.61 s) + ½ (−9.81 m/s 2 )(0.61 s) 2 = 1.83 md DOWN = d UP + ht of release = 1.83 m + 2.1 m = 3.93 mbut this is in the negative direction, so it is −3.93 mNow find t DOWN :−3.93 m = (0 m/s) t DOWN + ½ (−9.81 m/s 2 )(t DOWN ) 20.8012 s 2 = (t DOWN ) 2 , so t DOWN = 0.90 s5

Some Practice: The Shot PutNow calcuate t TOTt TOT = t UP + t DOWN = 0.61 s + 0.90 s = 1.51 sand finally, calculate the horizontal displacement:d H = v H t TOT = 10.4 m/s × 1.51 s = 15.7 mSo, now the shot travels 15.7 m horizontallyThe increased height of release resulted in a3.0 m (~20%) improvement in performance!6