{"id":3485,"date":"2023-10-05T10:42:25","date_gmt":"2023-10-05T10:42:25","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=3485"},"modified":"2023-10-05T10:42:53","modified_gmt":"2023-10-05T10:42:53","slug":"week-1-and-2-ss2-third-term-further-mathematics-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-1-and-2-ss2-third-term-further-mathematics-notes\/","title":{"rendered":"Week 1 and 2 – SS2 Third Term Further Mathematics Notes"},"content":{"rendered":"

THIRD TERM E-LEARNING NOTE
\n<\/strong>
\n\u00a0SUBJECT: FURTHER MATHEMATICS\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0CLASS: SS 2
\n<\/strong>
\n\u00a0SCHEME OF WORK
\n<\/strong>
\n\u00a0<\/p>\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
WEEK<\/strong><\/td>\n TOPIC<\/strong><\/td>\n<\/tr>\n
1.<\/td>\nRevision of Second Term Examination Questions.<\/td>\n<\/tr>\n
2. <\/td>\nProjectile: Trajectory of Projectile, Greatest Height Reached, Time of Flight, Range, and Projectile along Inclined Plane. <\/td>\n<\/tr>\n
3. <\/td>\nBinomial Expansion: Pascal Triangle Binomial Theorem of Negative, Positive and Fractional Power<\/td>\n<\/tr>\n
4. <\/td>\nMechanics: Vectors in Two and Three Dimension. Scalar Product of Vectors inThree Dimension.<\/td>\n<\/tr>\n
5<\/td>\nVector or Cross Product on Three Dimension.Application of Cross Product Cross Product of Two Vectors.<\/td>\n<\/tr>\n
6<\/td>\nReview of the Half Term Work. <\/td>\n<\/tr>\n
7. <\/td>\nIntegration: Indefinite Integrals Concept, Different Methods of Integration.e.g (Algebraic and Trigonometric Substitution by Parts and Partial Fractions.<\/td>\n<\/tr>\n
8. <\/td>\nIntegration Continued: Definite Integral, Area Under Curve.<\/td>\n<\/tr>\n
9. <\/td>\nIntegration Continued: Application of Integration to Kinematics Volumes of Solids at Revolution and Trapezium Rule.<\/td>\n<\/tr>\n
10. <\/td>\nCorrelation and Regression: Concept, Scatter Diagram, Regression Line, Coefficient of Regression, Rank Correlation and Product Moment Correlation Coefficient.<\/td>\n<\/tr>\n
11.<\/td>\nRevision.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

\u00a0REFERENCES
\n<\/strong>Further Mathematics Project 2 and 3.
\n\t\t\t<\/strong>
\n\u00a0
\n\u00a0WEEK ONE
\n\t\t\t\t<\/strong>REVISION OF SECOND TERM EXAMINATION QUESTIONS.<\/p>\n

\u00a0
\n\u00a0WEEK TWO
\n<\/strong>TOPIC:PROJECTILES: <\/strong>MOTION UNDER GRAVITY IN TWO DIMENSION,DERIVATION AND APPLICATION OF EQUATIONS INVOLVING GREATEST HEIGHT, TIME OF FLIGHT AND RANGE\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0
\n<\/strong>Motion Under Gravity in Two Dimensions:
\n<\/strong>If a particle is projected with an initial velocity u at angle to the horizontal, the prativle will be resolved into vertical and horizontal components of the velocity.
\n\"\"\/VY<\/sub><\/p>\n

\u00a0\"\"\/<\/p>\n

\u00a0
\n\u00a0
\n\u00a0Horizontal components: Vx = ucos
\n<\/strong>Horizontal distance: Sx = utcos
\n<\/strong>Vertical components: Vy = usin
\n<\/strong>Vertical distance: sy = utsin – \u00bd gt2 <\/sup>
\n\t\t\t<\/strong>Magnitude of the velocity, v = vx <\/sub>2<\/sup> + vy<\/sub>2<\/sup>
\n\t\t\t<\/strong>The acceleration due to gravity acts against the motion of the body, hence it is negative.
\nExample:
\n<\/em><\/strong><\/p>\n

    \n
  1. \n
    A particle is projected with an initial velocity of 46m\/s at an angle of 550<\/sup> to the horizontal. After 3 seconds, find: (i) the vertical component of the velocity (ii) horizontal component (iii) vertical distance traveled. (iv) Magnitude of the velocity.\n<\/div>\n

    SOLUTION:
    \n<\/em><\/strong> = <\/em>550 <\/sup>u=46m\/s
    \n (i) Vy = usin- gt
    \n<\/strong>\u00a0\u00a0\u00a0\u00a0 = 46 sin 55 \u2013 10 x 3
    \n = 37.68 \u2013 30
    \n = 7.68\/s
    \n(ii) Vx = ucos
    \n<\/strong> = 46 cos 55
    \n =26.38m\/s
    \n(iii) sy = utsin – (10×9)
    \n<\/strong> = 138sin55 \u2013 5×9
    \n = 113.04 \u2013 45
    \n = 68.04m
    \n(iv) = vx2 <\/sup>+vy2
    \n<\/sup>7.682<\/sup> + 26.382 <\/sup>= 58.98 + 695.9
    \n = 27.48m\/s
    \nEVALUATION: <\/em><\/strong>A particles is fired with an initial speed of 40m\/s at an angle of 300<\/sup> to the horizontal. Determine the vertical and horizontal components of the velocity after 2.5 seconds.<\/p>\n

    \u00a0GREATEST HEIGHT REACHED: <\/strong> when a projected particle reaches its greatest height, the vertical components become zero. Therefore;
    \nRecallVy, = usin \u2013 gt
    \n<\/strong>Squaring both sides, (Vy) 2<\/sup> = (usinn \u2013 gt)2
    \n<\/sup>Vy2 =<\/sup> u2<\/sup>sin2<\/sup> – 2gsy<\/sub>
    \n\t\t\t\tSince: vy =0, hence, 0 = u2<\/sup>sin2<\/sup> – 2gsy
    \n<\/sub>\"\"\/Sy<\/sub> =u2<\/sup>sin2<\/sup>
    \n\t\t\t\t 2g
    \nTherefore the greatest height reaches is represented by H=u2<\/sup>sin2<\/sup>
    \n\t\t\t\t2g
    \nTime taken to reach the greatest height<\/em><\/strong>: The time taken to reach the maximum height I at the point when the vertical component is zero. Hence,
    \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/em>,Vy = usin\u2013gt
    \n<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 0 = usin\u2013gt
    \n<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0T=usin<\/strong>
    \n\t\t\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0g
    \n<\/strong>
    \n\u00a0
    \n\u00a0Example:<\/em>
    \n\t\t\t\t\t<\/strong><\/p>\n

      \n
    1. \n
      A particle is projected with velocity 56m\/s at an angle of 600<\/sup> from a point O on a horizontal plane. The particle moves freely under gravity and hits the plain again A. Calculate, correct to 3 significant figures: (a) the greatest height above OA attainedby the particle (b) the time taken by the particle to reach A from O.\n<\/div>\n

      Solution:
      \n U = 56m\/s = 600
      \n<\/sup><\/li>\n<\/ol>\n

        \n
      1. \n
        Greatest height reached, h = U2<\/sup>sin2<\/sup>O<\/strong>\n\t\t\t\t\t\t<\/div>\n

        2g
        \n<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/strong>h = 562 <\/sup>x (sin 60)2<\/sup>
        \n\t\t\t\t\t\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2 x 9.8
        \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0h = 2352\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0h = 120m.
        \n\t\t\t\t\t\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 19.6
        \n(a) Time taken to reach A from O; t = usin
        \n<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0g
        \n<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/strong>t = 56 sin 60
        \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 9.8
        \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0T = 4.9secs.\n<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n

        Evaluation: A project is fired with a velocity of 45m\/s and at angle of elevation of 810<\/sup> to the horizontal. Find the time taken by the particle to reach its destination. (Take g = 10m\/s2<\/sup>)
        \nTime of flight:<\/em><\/strong> This is the time taken by a particle which is projected to return to its original point of projection. At this point the vertical distance becomes zero. Hemce,
        \n T = 2usin
        \n<\/strong>g
        \n<\/strong>Range: <\/strong>This is the horizontal distance covered when the particle returns to its original point of projection. The range is equal to the product of the horizontal component and the time of flight.
        \nHence,
        \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0R =ucos<\/strong>x2usin
        \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/strong>g
        \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0R =u2<\/sup> x 2sincos\u00a0\u00a0\u00a0\u00a0 (but; 2sincos = sin 2)
        \ng
        \nR = u2<\/sup> x 2sin
        \n<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0g
        \n<\/strong>Maximum range: <\/strong> A particle will cover a maximum range if it is projected at angle 450<\/sup> to the horizontal. That is; = 450<\/sup>. Thus sin2 =1
        \nHence, Rmax<\/sub>= U2<\/sup>
        \n\t\t\t\t<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0g
        \n<\/strong>Example:<\/strong> The vertical and horizontal components of the initial velocity of a projectile are 36m\/s and 64m\/s. find (i) initial velocity of the projectile (ii) the inclination to the horizontal at which the projectile was fired. (iii) the greatest height reached; (iv) the time of flight; (v) the horizontal range of the projectile.
        \nSolution:
        \n<\/strong>\u00a0\u00a0\u00a0\u00a0Vy<\/sub> = 36m\/s Vx<\/sub>= 64m\/s<\/p>\n

          \n
        1. \n
          VVx<\/sub>2 <\/sup> +Vy<\/sub>2<\/sup>\n\t\t\t\t<\/div>\n

          U = 642 + <\/sup>362<\/sup>; U = 73.43m\/s<\/p>\n

            \n
          1. \n
            Inclination to the horizontal; ( the angle of projection)\n<\/div>\n

            Vx = u cos
            \n<\/strong>64 = 73.43 cos
            \n<\/strong> = cos-1<\/sup> (64\/73.43); = 29.40<\/sup>
            \n\t\t\t\t\t\t\t<\/strong><\/li>\n<\/ol>\n<\/li>\n

          2. \n
            Greatest height reached; h = U2 <\/sup>sin
            \n\t\t\t\t\t<\/strong><\/div>\n

            \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2g
            \n<\/strong> h = 73.432<\/sup> x (sin 29.4)2<\/sup>
            \n\t\t\t\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02 x10
            \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0h = 5391.96 x 0.2410
            \n\t\t\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a020
            \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0h = 64.97m\n<\/li>\n

          3. \n
            Time of flight: T<\/strong> = 2usin<\/strong>\u00a0\u00a0\u00a0\u00a0\n<\/div>\n

            g
            \n<\/strong>\u00a0\u00a0\u00a0\u00a0<\/strong>T = 2x 73.43 x sin 29.4
            \n10
            \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 T = 7.2 secs.\n<\/li>\n

          4. \n
            Horizontal range: R = u2<\/sup> sin2<\/strong>\n\t\t\t\t<\/div>\n<\/li>\n<\/ol>\n

            g
            \n<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0R = 73.432<\/sup> x sin (2×29.4)
            \n\u00a0\u00a0\u00a0\u00a010
            \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0R = 461.2m
            \nEVALUATION:<\/strong>A particle is projected into the air with a speed of 50m\/s at an inclination sin-1<\/sup>(3\/5). Find the: (greatest height reached by the particles; (ii) horizontal range; (iii) time of flight<\/p>\n

            \u00a0Reading Assignment
            \n<\/strong>New Further Maths Project 2 page 262 -270.<\/p>\n

            \u00a0GENERAL EVALUATION
            \n<\/strong>1) A particle is projected with an initial speed of 45m\/s at an angle of 35 to the horizontal, find the time it takes for the particle to (i) reach the highest level (ii) return to its original level
            \n2) A particle is projected horizontally with a velocity of 40m\/s from the top of a tower 80.5m above the level ground find how far from the bottom of the tower the particle when it hits the ground
            \n3) A particle is projected into the air with a speed of 20m\/s at an inclination 30 to the horizontal , find the (i) greatest height reached (ii) horizontal range (iii) time of flight
            \n4) Show that a particle which is projected with a given velocity reaches its maximum range at an elevation of sin-1<\/sup> (21\/2<\/sup> \/2)<\/p>\n

            \u00a0WEEKEND ASSIGNMENT
            \n<\/strong>The vertical and horizontal components of the initial velocity of a projectile are 36m\/s and 64m\/s respectively find the
            \n1) greatest height reached a) 32.4m b) 97.2m c) 64.8m d) 16.2m
            \n2) time of flight a) 7.2s b) 3.6s c) 1.8s d) 14.4s
            \n3) horizontal range a) 23.04m b) 46.08m c) 11.5m d) 92.16m
            \n4) initial velocity of the projectile a) 73.4m\/s b) 146.8m\/s c) 36.7m\/s d)18.4m\/s
            \n5) inclination to the horizontal a) 19 b) 21 c) 29 d) 49<\/p>\n

            \u00a0THEORY
            \n<\/strong>1) Find the initial speed which a projectile must be subjected to give a maximum horizontal range of 490m
            \n2) Prove that the maximum range on a horizontal plane of a particle fired with velocity V at an angle x to the horizontal is V2<\/sup> \/ g<\/p>\n

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

            THIRD TERM E-LEARNING NOTE \u00a0SUBJECT: FURTHER MATHEMATICS\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0CLASS: SS 2 \u00a0SCHEME OF WORK \u00a0 WEEK TOPIC…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,275],"tags":[],"class_list":["post-3485","post","type-post","status-publish","format-standard","hentry","category-posts","category-third-term-ss2-further-mathematics"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3485","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=3485"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3485\/revisions"}],"predecessor-version":[{"id":3486,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3485\/revisions\/3486"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=3485"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=3485"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=3485"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}