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2 Analysing Projectile Motion Inestigate and analyse te motion of projectiles near te Eart s surface including a qualitatie description of te effect of air resistance; Let us break projectile motion into two cases; Case - off a orizontal cliff face and Case launced off te ground at an angle Case : Horizontal Projectile Motion 8 m/s 0 m Once te ball leaes te cliff te only force tat will act on it is te force of graity. Normally we break te motion into two components: te orizontal component and te ertical component Horizontal Component Since no force acts in te orizontal direction tis elocity will remain te same, namely 8m/s in te orizontal direction u at u at since tere is no acceleration in te orizontal direction we ae Vertical Component u 8ms Once it leaes te cliff it begins to experience te force of graity and its starts to ae a ertical elocity downwards We can work out te alue of te ertical elocity simply by using te equations of constant acceleration u at
3 Now we can add a subscript to sow tat we are looking at te component in te ertical direction u at So calling down as positie we can find te elocity in te ertical direction at any gien time, as it will be gien by te equation aboe But we will need to know ow long it will be in te air before it its te ground. So we use te equation of constant acceleration s ut at s ut at Notice we ae used te expression s wic means te displacement in te ertical direction. In te problem aboe te ertical displacement is 0metres and since it just leaes te cliff te initial ertical elocity, u is initially 0 Tere we get te following equation s ut at 0 at 0 0 t 0 5t t t t
4 3 Now we know ow long te ball stays in te air, namely seconds so we can find te final ertical elocity just before it its te ground u at ms downwards Now wat is te elocity on impact wit te ground? Since elocity is a ector we need to add te two elocities; te orizontal elocity and te ertical elocity 8 m/s.54 m/s 0 m/s So using Pytagoras we can find te elocity on impact c a b c c c c.54ms But wat about ow far it goes in te orizontal direction, tis is called te range To find te range we simply multiply te orizontal ertical wit te time it is in te air Now if we would use te equation of motion we would get te following equation
5 4 s ut at s ut at Since tere is no force in te orizontal direction, terefore tere is no acceleration in te orizontal direction our equation becomes te following Summary for doing problems inoling case s s s u t 8 6m Horizontal component Vertical Component elocity does not cange-u Vertical elocity does cange -U Range= u t To find te time we need to use te equations of constant acceleration and make sure we pick down as positie s ut at To find te final elocity we need to add ectorially te orizontal component wit te ertical component Try a few questions now on tis type of projectile motion 33 m m/s Find te following: a) Find te time it takes to reac te ground b) Find te orizontal elocity at t = 0.3 sec c) Find te ertical elocity at t = 0.3 sec d) Find te resultant elocity at t = 0.3 sec e) Find te resultant elocity just before it its ground.
6 5 3 m 7 m/s Find te following: f) Find te time it takes to reac te ground g) Find te orizontal elocity at t = 0.3 sec ) Find te ertical elocity at t = 0.3 sec i) Find te resultant elocity at t = 0.3 sec j) Find te resultant elocity just before it its ground. Case Case is a little more inoled Look at te diagram below 00 m/s 50º
7 6 Once again we break te problem into two components te orizontal component and te ertical component Initial Horizontal Component Initial Vertical Component 00m/s 00m/s U 50º 50º U U 00cos50 U 00sin50 Now tat we ae broken te problem into te two components we take eac component one at a time and find te releant information Horizontal Component - Te initial orizontal component will not cange proided te missile in on te same leel ( most of te time) - No force acting in te orizontal direction, terefore no acceleration in te orizontal direction Vertical Component - Tis component canges as tere is a force acting in te ertical direction, te force of graity - If we pick up to be positie ten te acceleration will be negatie - At te igest point ertically tere will be no more ertical elocity ( except for te orizontal elocity), so we can use te equation of constant acceleration to sole tis problem - Wen te object returns to te ground, te ertical displacement will be zero Lets us terefore analyse te question to see wat we can learn Question : Wat is te initial ertical elocity of te missile? : Since we ae already done it from te preious page te initial ertical elocity U 00sin ms Question : Wat is te initial orizontal l elocity of te missile?
8 7 U 00cos ms No remember tis orizontal elocity will remain constant for te entire question proided it stays on te same orizontal leel Question 3:Wat is te ertical elocity after second? Since we are asked to find te ertical elocity we will need to use ertical components. Remember we ae cosen to use up as being positie so we need to adjust te equations accordingly u at Question 4: Wat is te orizontal elocity after second? Te orizontal elocity does not cange terefore it is 8.56 m/s Question 5: Wat is te elocity after second? Tis is a tricky question since it is asking us to find te elocity after second, meaning we will need to add te two ectors, i.e te orizontal elocity and te ertical elocity and ten using Pytagoras find te resultant elocity m/s º 8.56 Te answer is 9.45 m/s in te direction gien by te two ectors Question 6: How ig will te missile go? Once again we are dealing wit ertical components, so we will need to use ertical elocities and let us not forget te initial direction-(wic is upwards as being positie) At te top of te eigt, te ertical elocity will be zero, so we can use te following equation u as
9 8 u as u as s 0s s s Question 7: How long does it take to get to te maximum eigt? We can use many different approaces but let us use a ery simple metod. At te top of te objects eigt, te ertical elocity will be 0 and we can ten use one of te simple equations to find te time it takes u at u at t 53. t So it takes just a bit oer 5 seconds or to be precise 5.3 seconds. Question 8: How long is te missile in te air? Ignoring air resistance, and looking at te symmetry of te pat te answer is twice te time to get to te igest point. so it will be seconds in te air. We could ae used anoter metod but tis is one of te simplest ways of soling it. Question 9: Find te final elocity it its te ground wit? Ignoring air resistance it sould it te ground wit te initial elocity it was projected i.e 00m/s Anoter way Find te final ertical elocity using u at and ten add te orizontal elocity wit tis to find te final elocity
10 9 Questions to try now 40 m/s 30º a) How far will tis object stay in te air for? b) How far will it go? ( its range) c) Find te initial ertical elocity of te object? d) Find te initial orizontal elocity of te object? e) How long does it take to get to te maximum eigt? f) Wat is its maximum eigt? g) Wat is te kinetic energy at te maximum eigt? ) Wat is te resultant elocity just before it its te ground? i) After 3 seconds ow ig is te object ertically? a. Wat is te objects orizontal distance from te start at 3 seconds? 330 m/s 40º a) How far will tis object stay in te air for? b) How far will it go? ( its range) c) Find te initial ertical elocity of te object? d) Find te initial orizontal elocity of te object? e) How long does it take to get to te maximum eigt? f) Wat is its maximum eigt? g) Wat is te kinetic energy at te maximum eigt? ) Wat is te resultant elocity just before it its te ground? i) After 3 seconds ow ig is te object ertically?
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