{"id":2994,"date":"2023-10-04T08:58:33","date_gmt":"2023-10-04T08:58:33","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=2994"},"modified":"2023-10-04T09:04:49","modified_gmt":"2023-10-04T09:04:49","slug":"week-4-ss2-first-term-physics-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-4-ss2-first-term-physics-notes\/","title":{"rendered":"Week 4 – SS2 First Term Physics Notes"},"content":{"rendered":"

WEEK 4
\n<\/strong><\/p>\n

PROJECTILES AND ITS APPLICATION
\n<\/h2>\n

CONTENTS
\n<\/h2>\n
    \n
  • \n
    Meaning of projectile\n<\/div>\n<\/li>\n
  • \n
    Terms associated with projectiles\n<\/div>\n<\/li>\n
  • \n
    Uses of projectile\n<\/div>\n<\/li>\n<\/ul>\n

    MEANING OF PROJECTILE<\/strong>
    \n\t\tA projectile motion is one that follows a curved or parabolic path .It is due to two independent motions at right angle to each other .These motions are <\/p>\n

      \n
    1. \n
      a horizontal constant velocity\n<\/div>\n<\/li>\n
    2. \n
      a vertical free fall due to gravity\n<\/div>\n<\/li>\n<\/ol>\n

      Examples of projectile motion are the motion of;<\/p>\n

        \n
      1. \n
        a thrown rubber ball re-bouncing from a wall\n<\/div>\n<\/li>\n
      2. \n
        An athlete doing the high jump\n<\/div>\n<\/li>\n
      3. \n
        A stone released from a catapult\n<\/div>\n<\/li>\n
      4. \n
        A bullet fired from a gum\n<\/div>\n<\/li>\n
      5. \n
        A cricket ball thrown against a vertical wall.\n<\/div>\n<\/li>\n<\/ol>\n

        \"\"\/\"\"\/
        \n\t\t U y
        \n\t\t\t<\/strong>
        \n\t\t\"\"\/
        \n\t\t\"\"\/\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0Hmax<\/sub>
        \n\t\t \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0t
        \n\u00a0
        \n\u00a0\u00a0
        \n\u00a0 )\u03b8
        \n\"\"\/P\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0Ux\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/sub>Q
        \nUy<\/sub> = U sin \u03b8\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(vertical component) \u00a0\u00a0\u00a0\u00a0——————- 1
        \nUx<\/sub> = U cos \u03b8\u00a0\u00a0\u00a0\u00a0(horizontal component)\u00a0\u00a0\u00a0\u00a0——————- 2<\/p>\n

        TERMS ASSOCIATED WITH PROJECTILE
        \n\t\t\t<\/h2>\n
          \n
        1. \n
          Time of flight (T) <\/em>– The time of flight of a projectile is the time required for it to return to the same level from which it projected.\n<\/div>\n<\/li>\n<\/ol>\n

          t= time to reach the greatest height
          \nV = u + at\u00a0\u00a0\u00a0\u00a0(but, v =o, a = -g)
          \n\u03b8= u sin \u2013 gt
          \nt = U sin \u03b8\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0——————- 3
          \ng
          \nT = 2t = 2U sin \u03b8\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0——————- 4
          \ng
          \n2.\u00a0\u00a0\u00a0\u00a0The maximum height (H)<\/em> – is defined as the highest vertical distance reached measured from the horizontal projection plane.
          \nFor maximum height H,
          \nV2<\/sup> = U2<\/sup> sin2<\/sup>\u03b8 – 2g H
          \nAt maximum height H, V=0
          \n2gH = U2<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0——————- 5
          \n 2g
          \n3.\u00a0\u00a0\u00a0\u00a0The range (R)<\/em> – is the horizontal distance from the point of projection of a particle to the point where the particle hit the projection plane again.
          \nHorizontally, considering the range covered
          \nUsing S= ut + \u00bdat2 <\/sup>(where a=0 for the horizontal motion)
          \nOR
          \nS = R = U cos\u03b8 x t (distance = velocity x time; there time is the time of flight)<\/p>\n

          \u00a0R = U cos \u03b8 (2 U sin \u03b8)
          \ng
          \nR = 2U2<\/sup> sin \u03b8 cos \u03b8
          \ng
          \nFrom Trigonometry function
          \n2 sin \u03b8 cos \u03b8 = sin 2\u03b8
          \nR= U2<\/sup> sin 2\u03b8
          \n\t\tg
          \nFor maximum range \u03b8 = 450<\/sup>
          \n\t\tSin2\u03b8 = sin 2 (45) = sin 900<\/sup> = 1
          \nR= U2<\/sup>
          \n\t\t g
          \nRmax<\/sub> = U2<\/sup>
          \n\t\tg
          \n\u00a0
          \n\u00a0<\/p>\n

          USE OF PROJECTILES
          \n<\/h2>\n

          1.\u00a0\u00a0\u00a0 To launch missiles in modern warfare
          \n2.\u00a0\u00a0\u00a0 To give athletes maximum takeoff speed at meets
          \nIn artillery warfare, in order to strike a specified target, the bomb must be released when the target appears at the angle of depression \u03c6 given by:
          \nTan \u03c6 =1\/u \u221agh\/2 \u00a0
          \nEXAMPLES
          \n1.\u00a0\u00a0\u00a0 A bomber on a military mission is flying horizontally at a height of zoom above the ground at 60kmmin-1<\/sup>. lt drops a bomb on a target on the ground. Determine the acute angle between the vertical and the line joining the bomber and the tangent at the instant. The bomb is released<\/p>\n

          \u00a0\"\"\/\"\"\/\"\"\/Ux 60m\/ min
          \n\u00a0
          \n\u00a0\u00a0
          \n\u00a0\u00a0
          \n\u00a0 3,000m
          \n\"\"\/\u00a0
          \nHorizontal velocity of bomber = 60km\/min= 103 <\/sup>ms-1<\/sup>
          \n\t\tBomb falls with a vertical acceleration of g = 10m\/s
          \nAt the release of the bomb, it moves with a horizontal velocity equals that of the aircraft i.e. 1000m\/s
          \nConsidering the vertical motion of the bomb we have
          \nh =ut+1\/2 gt2 <\/sup>(u=o)
          \nh =1\/2gt2<\/sup>
          \n\t\tWhere: t is the time the bomb takes to reach the ground: 300=1\/2gt2<\/sup>
          \n\t\tt2<\/sup>= 600
          \nt=10\u221a6 sec
          \nConsidering the horizontal motion we have that horizontal distance moved by the bomb in time t is given by
          \ns =horizontal velocity x time
          \ns = 1000 x10\u221a6
          \ns = 2.449×104<\/sup> m
          \nBut tan\u03b8 = s = 2.449 x 104<\/sup>
          \n\t\t3,000 3,000
          \n\u00a0 \u03b8 =83.020<\/sup>\u00a0
          \n2. A stone is shot out from a catapult with an initial velocity of 30m at an elevation of 600<\/sup>. Find
          \na.\u00a0\u00a0\u00a0\u00a0the time of flight
          \nb.\u00a0\u00a0\u00a0\u00a0the maximum height attained
          \nc.\u00a0\u00a0\u00a0\u00a0the range
          \na.\u00a0\u00a0\u00a0\u00a0The time of flight
          \nT = 2U sin \u03b8
          \n\t\tg
          \nT= 2 x 30 sin 600<\/sup>
          \n\t\t10
          \nT= 5.2s
          \nb.\u00a0\u00a0\u00a0\u00a0The maximum height,
          \nH=U2<\/sup> sin2<\/sup> \u03b8
          \n2g
          \nH = 302<\/sup> sin2<\/sup> (60)
          \n20
          \n\t\t\tH = 33.75 m<\/p>\n

          \u00a0 c.\u00a0\u00a0\u00a0\u00a0The range,
          \nR = U2<\/sup>sin 2\u03b8
          \n\t\tg
          \nR = 302<\/sup> sin 2<\/sup> (60)
          \n\t\t10
          \nR = 90 sin 120
          \nR = 77.9 m<\/p>\n

          \u00a0
          \n\u00a0
          \n\u00a03. A body is projected horizontally with a velocity of 60m\/s from the top of a mast 120m above the grand, calculate
          \n(i) Time of flight, and (ii) Range
          \n\"\"\/\"\"\/\"\"\/\u00a0
          \n\"\"\/\"\"\/ 60 m\/s
          \n\u00a0
          \n\u00a0 120
          \n\"\"\/
          \n\t\t R<\/p>\n

            \n
          1. \n
            s =ut+1\/2gt2<\/sup>\n\t\t\t\t<\/div>\n<\/li>\n<\/ol>\n

            a=g, u=0
            \n120= \u00bd (10)t2<\/sup>
            \n\t\tt2<\/sup> = 24
            \nt = 24
            \nt = 4.9s<\/p>\n

              \n
            1. \n
              Range =u cos\u03b8 x T.\n<\/div>\n<\/li>\n<\/ol>\n

              But in this case \u03b8 = 0
              \nCos 0 =1
              \nR = ut
              \nR = 60x 4.9
              \nR =294m
              \n4. \u00a0\u00a0\u00a0\u00a0A stone is projected horizontally with a speed of 10m\/s from the top of a tower 50m high and with what speed does the stone strike the ground?
              \nSolution<\/strong>
              \n\t\tv2<\/sup>=u2<\/sup> + 2gh
              \nv2<\/sup>=102<\/sup>+ (2x10x50)
              \nv2<\/sup>=100+1000
              \nv2<\/sup>=1100
              \nv2<\/sup>=33.17m\/s<\/p>\n

              \u00a0<\/p>\n

                \n
              1. \n
                A projectile is fired at an angle of 60 with the horizontal with an initial velocity of 80m\/s. Calculate:\n<\/div>\n
                  \n
                1. \n
                  the time of flight\n<\/div>\n<\/li>\n
                2. \n
                  the maximum height attained and the time taken to reach the height\n<\/div>\n<\/li>\n
                3. \n
                  the velocity of projection 2 seconds after being fired (g = 10m\/s)\n<\/div>\n<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n

                  \u03b8 =60; u =80m\/s<\/p>\n

                    \n
                  1. \n
                    T = 2 U sin \u03b8\n\t\t\t\t<\/div>\n

                    g\n<\/li>\n<\/ol>\n

                    T = 2×80 sin 60
                    \n\t\t10
                    \nT = 13 .86 s<\/p>\n

                      \n
                    1. \n
                      A.\u00a0\u00a0\u00a0\u00a0H = u2<\/sup> sin 2\u03b8\n\t\t\t\t<\/div>\n

                      2g
                      \nH = 80 x 80 x sin60
                      \n\t\t\t\t20
                      \nH = 240 m\n<\/li>\n<\/ol>\n

                      B.\u00a0\u00a0\u00a0\u00a0t = U sin \u03b8
                      \n\t\tg
                      \nt = 80 sin 60
                      \n\t\t10
                      \nt = 6.93 s
                      \n\u00a0
                      \n\u00a0R =U2<\/sup> sin 2 \u03b8
                      \n g
                      \nR = 802<\/sup>sin2 (60)
                      \n 10
                      \nR = 640 sin 120
                      \nR = 554.3m
                      \n\u00a0
                      \n\u00a0<\/p>\n

                        \n
                      1. \n
                        Vy<\/sub> = U sin \u03b8 \u2013 gt\n<\/div>\n<\/li>\n<\/ol>\n

                        Vy<\/sub> = 80 sin 60 \u2013 20
                        \nVy<\/sub> = 49.28m\/s
                        \n\t\tUx<\/sub> = U cos \u03b8
                        \nUx<\/sub> = 80 cos 60
                        \nUx<\/sub> = 40 m\/s
                        \nU2<\/sup> = U2<\/sup>y + U2<\/sup>x
                        \nU2<\/sup> = 49.282<\/sup> + 402 <\/sup>
                        \n\t\tU = \u221a (1600+ 2420)
                        \nU = 63.41 m\/s
                        \n\u00a0
                        \n\u00a0CLASSWORK
                        \n<\/strong><\/p>\n

                          \n
                        1. \n
                          (a) Define the term projectile (b) mention two application of projectiles\n<\/div>\n<\/li>\n
                        2. \n
                          A ball is projected horizontally from the top of a hill with a velocity of 30m\/s. if it reaches the ground 5 seconds later, the height of the hill is\n<\/div>\n<\/li>\n
                        3. \n
                          A stone propelled from a catapult with a speed of 50m\/s attains a height of 100m. Calculate: a. the time of flight b. the angle of projection c. the range attained.\n<\/div>\n<\/li>\n<\/ol>\n

                          ASSIGNMENT
                          \n<\/strong>SECTION A
                          \n<\/strong><\/p>\n

                            \n
                          1. \n
                            A stone is projected at an angle 60 and an initial velocity of 20m\/s determine the time of flight (a) 34.6s (b) 3.46s (c) 1.73s (d) 17.3s (e) 6.92s\n<\/div>\n<\/li>\n
                          2. \n
                            A stone is projected at an angle 60 and an initial velocity of 20m\/s determine the time of flight (a) 34.6s (b) 3.46s (c) 1.73s (d) 17.3s (e) 6.92s\n<\/div>\n<\/li>\n
                          3. \n
                            For a projectile the maximum range is obtained when the angle of projection is; (a) 600<\/sup> (b) 300<\/sup> (c) 450<\/sup> (d) 900<\/sup>\n\t\t\t\t<\/div>\n<\/li>\n
                          4. \n
                            The maximum height of a projectile projected with an angle of to the horizontal and an initial velocity of U is given by\n<\/div>\n<\/li>\n<\/ol>\n

                            (a) U sin2<\/sup> \u03b8 (b) U2<\/sup> sin2<\/sup>\u03b8 (c) U2<\/sup>sin \u03b8 (d) 2U2<\/sup>sin2<\/sup>\u03b8
                            \n\t\tg\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02g\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0g\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0g
                            \nUse this information to answer questions 5 and 6: An arrow is shot into space with a speed of 125m\/s at an angle of 150<\/sup> to the level ground. Calculate the:<\/p>\n

                              \n
                            1. \n
                              Time of flight (a) 5seconds (b) 6.47seconds (c) 16.01seconds (d) 4.7seconds\n<\/div>\n<\/li>\n
                            2. \n
                              Range of the arrow (a) 350m (b) 781.25m (c) 900m (d) 250.71\n<\/div>\n<\/li>\n<\/ol>\n

                              SECTION B\u00a0
                              \n<\/strong><\/p>\n

                                \n
                              1. \n
                                A gun fires a shell at an angle of elevation of 300<\/sup> with a velocity of 2x10m. What are the horizontal and vertical components of the velocity? What is the range of the shell? How high will it rise?\n<\/div>\n<\/li>\n
                              2. \n
                                (a) What is meant by the range of a projectile? (b) An object is projected into the air with a speed of 50m\/s at an angle of 300<\/sup> above the ground level. Calculate the maximum height attained by the object\n<\/div>\n<\/li>\n<\/ol>\n

                                \u00a0
                                \n\u00a0
                                \n\u00a0
                                \n\u00a0<\/p>\n","protected":false},"excerpt":{"rendered":"

                                WEEK 4 PROJECTILES AND ITS APPLICATION CONTENTS Meaning of projectile Terms associated with projectiles Uses…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,242],"tags":[],"class_list":["post-2994","post","type-post","status-publish","format-standard","hentry","category-posts","category-first-term-ss2-physics"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2994","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/comments?post=2994"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2994\/revisions"}],"predecessor-version":[{"id":2995,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2994\/revisions\/2995"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=2994"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=2994"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=2994"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}