physics-notes-motion-in-a-plane (1)

1. MOTION IN A PLANE

2. WŚLJƐiĐƐ^ŵĂƌƚBŽŽŬlĞƚ MOTION IN A PLANE Motion with uniform acceleration in a plane Consider a point object moving in XY plane with an uniform acceleration . Let us suppose ‘O’ be the origin for measuring time and position of the object. Let the object be at positions A and B at times t1 and t2 respectively, where OA 1 a  r2. r and OB LJ B A dž K v1 and v2 Let be the velocities of object at instants t1 and t2 respectively, then constant acceleration is given by  v2  v1 t2  t1 a v2 1  2  t1 ) v a(t  v a(t  1 t1 ) v2 2 If t1 = 0, t2 = t, v1 2 v u and v then  u at v t ation can be expressed in terms of rectangular component The above equ follows. ˆ u ˆ wheru  u2 u2 y j x y i n d n m f gul r mp n n s in XY plane as u  u i  x ere where v  ˆ v ˆ v  v i  x v2 v2 x y j y ˆ ˆ a2 a2 x  a i  x where a  y j a a y  x  x Also, vx = ux + axt and vy = uy + ayt  u t  1 a t2 x Displacement in x-direction is given by 0 x 2 And displacement in y-direction is given byy  y  u t  1 a t2 0 y y 2  Projectile Projectile is the name given to a body thrown with some initial velocity making an angle [ 90] with the horizontal direction, and then allowed to move in two dimensions under the action of gravity alone, without being propelled by any engine or fuel. ϯ

3. WŚLJƐiĐƐ^ŵĂƌƚBŽŽŬlĞƚ The path followed by a projectile is called its ‘trajectory’. Examples of projectile are: (i) A ball hit by a bat (ii) A bullet fired from a gun or pistol (iii) A javelin thrown by an athlete (iv) A shot-put sphere thrown by an athlete (v) A body dropped from an aeroplane in flight / bus / train In the above examples, we find that a projectile moves under the combined effect of two velocities: 1. A uniform velocity in the horizontal direction, which would not change provided there is no air resistance. 2. A uniformly changing velocity in the vertical direction due to gravity. To study the motion of a projectile, the following assumptions are made. (i) There is no esista c of a r o resistance of air. (ii) t due to rotation of earth and curvature of the earth is ne The effecdue to ro ati n f a th d ur of he earth is nglected. (iii) The acceleration du t grav ty (g s c ns t in gnitude and n direction at all eration due to gravity (g) is constant in magnitude and i otion of projectile. points of the m tion of proje tile  Horizontal Proj c i When a body is projec ed ho izontal y w h a v lo ity f om po t abo is called a “horizontal proje tile”. When a ston is projected horizontall ojectile projected horizontally with a velocity from a point abov ontal projectile”. When a stone is projected horizontally a tower of height ‘h’ it describes a parabolic path as sho from the top of to g h’ it describes a parabolic path as shown in figure. e the ground level, it with a velocity ‘u’ Ƶ WaƌaďŽlicƉaƚŚ aƌaďŽlicƚŚ Ś  Ƶ  Z  ǀ Őƚ  2h g (i) Time of descent, t  (independent of ‘u’) 2h g (ii) Horizontal displacement (or) range is R  u (iii) u2 2gh u2 g2t2 The speed with which it hits the ground is v  The angle at which it strikes the ground is  tan1 gt  tan1      (iv) 2gh u     u       ϰ

4. WŚLJƐiĐƐ^ŵĂƌƚBŽŽŬlĞƚ (v) If is angle of elevation of point of projection from the point where the body hits the gt2 ground, then tan  h  R ut 2u 2 2  gt tan  tan     Oblique Projectile LJ Ƶ ƵcŽƐ  ƵLJ H  WaƌaďŽla dž Ƶdž Z Any body projected into air with some velocity at an angle [90and 0] with the horizontal is called an “obliqueprojectile”. (i)Horizontal component of velocity is ux = u cos , remains constant throughout the journey. ponent of velocity ux= u sinvaries at the rate of ‘g’. (ii)Vertical component of locity u u s n , ari t the te of ‘g . (iii)After a time ‘t’: ‘t’: omponent of velocity is vx= u cos(= ux) (a)Horizontal c mponent of velocity s v o (= ) ponent of velocity is vy= uygt = u sin gt (b)Vertical component of v ocity s y y gt = u sin gt (c)Resultant velocity i v  v22 v ocity is x y    v v  (d)Direction of  x  y tan ta 1 velocity is given by elo ty by  ngle thatvmakes with horizontal. where is the angl t e with horizontal. (e)Horizontal displacement in a time ‘t’ is x = uxt = (u cos ) t (f)Vertical displacement in a time ‘t’ is y  u t  1 gt2 (u sin )t  1 gt2 y 2 2 (g)Net displacement of the body is S  x2 y2 (h)Equation of trajectory (which is a parabola) of an oblique projectile is g   2u cos   y  (tan )x  x2 Ax  Bx2  2 2  From the above equation, o= tan1 (A) oRange of projectile is R  A B A2 4B oMaximum height is H  ϱ

5. TO DOWNLOAD/VIEW FULL FILE Download Android App Fast Forward a work of Adhipati Creations that provides the best app for NEET, JEE, BITSAT, CUET and CBSE exam preparation. Visit Our Website :- https://efastforward.in