Motion in a Plane Notes - Projectile Motion, Study notes of Physics

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PROJECTILE MOTION PROJECTILE MOTION MOTION IN TWO DIMENSIONS DISPLACEMENT An object moving in a plane is said to have two dimensional motion. The two dimensional motion is equal to vector sum of two one dimensional motions along mutually perpendicular direction. Let the position of point P at a time t be given by position — P(x, y) vector OP=r rsin@ , + * + * r=ircos@ + jrsin@ = Ix + ]y x o Y x=rcos@ Let the position of point P at time t, be described by position vector r,=x,i + y,] and at time t, position Q is given by position vector 5 =X,i + Yo] Y (%y)P from AOPQ => OP + PQ = 00 or PQ =0Q- oP 4 + + + -. Displacement PQ = 4r =h-h in time interval dt = (t, - t,) or ér = (%) 14 Y2 I-04 1+y,]) = (x, —x,)i+ (V2 ys) =8xi+ dy j I bd 2% displacement along X-axis 8x displacement along Y-axis = ay = y, - y, Thus displacement in 2 dimensions is equal to vector sum of two one dimensional displacements along mutually perpendicular directions. + Let particle move with uniform velocity y at an angle 0 with X-axis. Then in component form v =v, i4 VI here v,=vcos 0 and My BM sin 0 and 6x = v, ot ; by = v, ot x ¥ or ax = (v cos 8) dt ; dy = (v sin 6) at so with v, we get displacement along X-axis only and vy gives displacement along ‘Y-axis only. PROJECTILE MOTION And if particle is moving with uniform acceleration a, then d=a,i+a,j If direction of a makes angle ¢ with X-axis then a, = a cos > and ayaa sin $ are components of a. Due to a,, there is change in X component of velocity only with no change in Y-component. Similarly a, will change only Y component of velocity at time t so v, =u, + a, t (here u, and u, are components of initial velocity) vw =u tat and v, =u, +a, t hence V,i+Vy] = (u, +a,t)i + (u, +a,t)j = . = = + = = (u,i+u,j)+(airayst => or ly=u+at and similarly component of displacement are 1 1 s=u,t+ 5 ah and s=utt 5 at ie 4 r +. 1 ? 7 hence 5S, 1+8,j= (u, i +U,))+ 5 (8x ita, jt + = iam or s=ut+—at 2 PROJECTILE MOTION Ex. A body which is in flight through the atmosphere but is not propelled by any fuel is called a projectile. A body or particle moving in atmosphere under effect of gravity only. Motion of projectile is two dimensional motion in a vertical plane. Stone thrown in air by a boy, Bullet fired from a gun, Javelin thrown by an athlete, Football kicked by a player, Bomb released from an aeroplane in flight. Trajectory : Path followed by a projectile is known as trajectory of projectile. When we consider motion of a projectile, following assumptions are made : (i) There is no resistance due to air. (ii) No effect due to curvature of earth. (iii) No effect due to rotation of earth. (iv) For all points on trajectory acc. due to gravity g (which is downward) remains same. Two types of projectile motion : (i) Horizontal projectile : Body projected horizontally from a certain height with a certain velocity. (ii) Oblique Projectile : Body projected at a certain angle with the horizontal. PROJECTILE MOTION PRINCIPLE OF PHYSICAL INDEPENDENCE OF MOTIONS Motion of projectile is two dimensional motion in a vertical plane. It can be resolved in two motions along horizontal & vertical direction These two motions are independent of each other. This is called principle of physical independence of motions. At any instant velocity of projectile has two components : (i) Horizontal Component : No acc. along horizontal (a, = 0) so velocity along horizontal remains unchanged throughout the flight. Horizontal motion is uniform motion. (ii) Vertical Component : Acceleration due to gravity in downward direction will change the vertical component of velocity continuously throughout the motion. Vertical motion is uniformly accelerated motion. OBLIQUE PROJECTILE MOTION Consider the motion of a bullet which is fired from a gun so that its initial velocity u makes an angle 6 with the horizontal direction. Let us take X-axis along ground and Y-axis along vertical. u can be resolved as u, = ucos @ (along horizontal) & u, =u sin 0 (along vertical) motion of bullet can be resolved into horizontal and vertical mation. (i) In horizontal direction there is no acc. so it moves with constant velocity v, = u, = u cos 6 So distance traversed in time t is x = u, t or x = (u cos @) tort = ucos@: ween fl) u,=ucos 6 ¥, iY | _ Poe | =e yo iy i 2 : usind Cc 4 ----2 : = ! i ~ 1 ucos6 \ fe ; x oe oop > aan i ic Horizontal Vertical component component The motion in the vertical direction is the same as that of a ball thrown upward with an initial velocity u, = usin @ and acc = — g (downward). So at time t vertical component of velocity v, = u, — gt = u sin 8 — gt cvveail) 1 Displacement along y direction y = (usin &#t— > gt? wee (lil) PROJECTILE MOTION Substituting the value of t from eqn. (i) in eqn. (iii) . / 1 x \ we get y=(usiné -= q eI ({—.) 30(5) Te =x tand—- or y 2u? cos*0 This is eqn. of parabola. The trajectory of projectile is parabolic The projectile will rise to maximum height H (where v, = u cos 0, v, = 0) and then move down again to reach the ground at a distance R from origin. Setting x=R and y=O (since projectile reaches ground again) g a: = = : R? O=R tan6é >u*c0s6 2 2, j We get R= 2u*cos70 7 sing g cos 2 or R= u - sindcosé 2 g B a Zeal uw or Range |R = u? sin 20 z tee g a RR sasrerensenererenernrerannn 3 of * If time for upward journey is t : at highest point v, = 0 so = 0 = (u sin 6) - gt (v, = u, — gt) te using or g T=2t (it will take same time for downward journey) r-aene Time of flight At the highest point y = H and v, = 0 5 Maximum Height and Time of So that H= Uy [v2 =u? -2gy] Flight Depends on Vertical 29 _- Component of Initial Velocity 2 ein 2 or fell vans Maximum Height 29 we can also determine R as follows x=ut so R=u,.T _ i [ zsh = (u cos 8) lg u? sin20 R= or g velocity at time t v,=v,, i+ Via j =(ucos6) | +(usiné—gt) j —— v =,/u*cos74 + (usin@—gt)? PROJECTILE MOTION Note : 2u?sinfcosé g (ii) Vertical component of velocity v, = 0, when particle is at the highest point of trajectory. 4 (i) Alternative eq". of trajectory y = x tan@ 1 “7 | where R= } (iii) Linear momentum at highest point = mu cos 4 is in horizontal direction. (iv) Vertical component of velocity is tive when particle is moving up. (v) Vertical component of velocity is -ive when particle is moving down. i> > Vv (vi) Resultant velocity of particle at time t v=,/v;+v, at an angle $= tan {=} # (vii) Displacement from origin, s = ,/x? + y? Special Points : (1) The three basic equation of motion, i.e. v=u+at s=ut+ 5 at? v7 = u? + 2as For projectile motion give : i Ze Zejpr2 7a using RU sin 26 Haw sin?6 g g 2g (2) In case of projectile motion, The horizontal component of velocity (u cos @), acceleration (g) and mechanical energy remains constant. Speed, velocity, vertical component of velocity (u sin 0), momentum, kinetic energy and potential energy all change. Velocity and K.E. are maximum at the point of projection, while minimum (but not zero) at the highest point. SAME RANGE (3) If angle of projection is changed from a+ p= 90° o—_,9° = (90 - 6), a and § are two angles of projection with same velocity. R= u’sin20° u?sin2(90—0) u’sin20 _ R then range g g g High So a projectile has same range for angles of projection @ and (90 — 4) But has different time of flight (T), maximum height (H) & trajectories u’cos2a Range is also same for 0, = 45° — a and 6, = 45° + a. equal = PROJECTILE MOTION (4) For maximum Range R=R,,, = 20 = 90° for 6 = 45° ue Rix = Oo [For sin 26 = 1 = sin 90° or 6 = 45°) When range is maximum = Then maximum height reached 2 u?sin?45 Hea - = (When R = H 29 (When R,,.,) or 4g hence maximum height reached (for R__,) i If a body is projected from a place above the surface of earth, then for the maximum range, the angle of projection should be slightly less than 45°. For javelin thraw and discus throw, the athlete throws the projectile at an angle slightly less than 45° to the horizontal to achieve the maximum range. (5) For height H to be maximum 2 ej? H=" = a =max i.e. sin?0 = 1 (max) or for 0 = 90° Es So that H,,,, = 2q When projected vertically (i.e. at @ = 90°) u’sin(2x90°) _u? sin180° _ g g u? u? A... = 29 (For vertical projection) and R_., = OQ (For oblique projection with same velocity) in this case Range R= 0 - Rinax so Hiax ~ 2 u? If a person can throw a projectile to a maximum distance (with 6 = 45°) R_ |. = o° t ' *, Rynax The maximum height to which he can throw the projectile (with 6 = 90°) H,.. = 2 (6) At highest point u’sin?0 1 Potential energy will be max and equal to (PE), = mgH = mg. or (PE), = 3 mu? sin? 0. While K.E. will be minimum (but not zero) and at the highest point as the vertical component of velocity is zero. 1 5 1 2 1 = —mv,=—m(ucos0 == a z (KE), ova ( ) > mu? cos 0 1 . 1 1 so (PE), + (KE), = mu sin? 6 + 2 mu? cos? @ = 2 mu? = Total M.E. So in projectile motion mechanical energy is conserved. PE 1 sin 20 Soif@=45° tan? @= 1 — | =e = 2 1 (Fe). 4 Hs 3 tan*@ PE=KE.= 2 M.E. at highest point ie. if a body is gm cos*0 projected aft an angle @ = 45° to the honzontal then at highest point, half of its MLE. is K.E. and half is PE. 6 PROJECTILE MOTION Ex. Sol. Ex. Sol. Sol. Ex. Sol. Sol. (7) In case of projectile motion if range R is n times the maximum height H, i.e. R = nH u?sin26 4 usin? 6 then g = 2g > 5 i nsing or cos 6 = 3 fA” or tano=4 => 0=tan {=| n ny (8) Weight of a body in projectile motion is zero as it is a freely falling body. A projectile of mass m is projected with velocity v at an angle 6 with the horizontal. What is the magnitude of the change in momentum of the projectile after time t ? Change in momentum = impulse = force = time = mgt. A projectile of mass m is fired with velocity v at an angle 6 with the horizontal. What is the change in momentum as it rises to the highest point of the trajectory? vsin 6 . mv sin @ Change in moment = force = time = mgx A ball of mass m is thrown vertically upwards. Another ball of mass 2 m is thrown up making an angle 0 with the vertical. Both of them stay in air for the same time. What is the ratio of their maximum heights? Since the two bodies are in air for equal interval of time therefore the velocity of projection of first body is equal to the vertical component of the velocity of projection of the second body. So, the maximum heights are the same. The required ratio is 1: 1. iy 2 v3 Vv What is the angle of projection of an oblique projectile if its range is 2g v® sin 20 V3 Comparing the given expression with gq? we get sin 26 = e or 26 = 60° or @ = 30° Two boys stationed at A and B fire bullets simultaneously at a bird stationed at C. The bullets are fired fromA and B at angles of 53° and 37° with the vertical. Both the bullets fire the bird simultaneously. What is the value of v, if v, = 60 units? Given : tan 37° = : The vertical components must be equal. Vv, COS 53° = V_, COS a cas 37° 7 Ya = Ye cos (90° — 37°) 60 or v, = 60 cot 37° = tan37° 60x4 ; = 3 = 80 units PROJECTILE MOTION HORIZONTAL PROJECTILE MOTION Suppose a body is thrown horizontally from point O, with velocity u. Height of O from ground = H. Let X-axis be along horizontal and Y-axis be vertically downwards and origin O is at point of projection as shown in fig. Let the particle be at P at a time t. The co-ordinates of P are (x, y). Distance travelled along X-axis at time t with uniform velocity o_UEy i.e. velocity of projection and without acceleration. The horizontal component of velocity v, = u and horizontal displacement x =u.t mm) displacement along vertical direction is y to calculate y, consider vertical motion of the projectile we den neeceeeneenne TT enccemenn cnn initial velocity in vertical direction u, = 0. acceleration along y direction a,=g (acc. due to gravity) So v,=a,t (y comp. of velocity at time t) , = gt seven (Ii) (as body were dropped from a height) or v Resultant velocity at time tis v= ui+ (gt)j v=4u* +(gt)’ t if f is the angle of velocity with X-axis (horizontal) tanp -< 1 d =~>gt? cove an y 39 (iii) =19(*) f tion (i) t= ~ or Y=5 ba [from equation (i) t = 7d 9 ye =x or ¥ Due g \ = 2 k=—~ or y = kx here ut (k is constant) This is eqn. of a parabola. A body thrown horizontally from a certain height above the ground follows a parabolic trajectory till it hits the ground. 1 (i) Time of flight |r - |20 las y= 5a", g _ [24 Ter (ii) Range = horizontal distance covered = R. R =u * time of flight [2H R=u- .jJ— “ = z a [y A Du? R?] (iii) Velocity when it hits the ground |v, = Ju? + 2gH PROJECTILE MOTION Ex. Sol. Ex. Sol. Ex. Sol. A projectile is fired with a horizontal velocity of 330 ms~' fram the top of a cliff 80 m high. How long will it take to strike the ground at the base of the cliff? With what velocity will it strike? Neglect air resistance. . oO u,= 330 m/s Let us consider the vertically downward motion. ‘u’=0, a = +9.8 mis?, $=80m t=?, . 4 1 Using S =ut + 2 at?, we get 80 = 2 x 9.8 160 or t= ign = 16.33 => t = 4.04 sec Distance from base R = ut = 330 * 4.04 = 1333.20 m. Now, v, =u, +a t= 9.8 x 4.04 ms‘ = 39.59 m/s Speed = 3307 +39.59* = 332.37 m/s 39.59 tan# =——-_ = 0.12 = ° and anp 330 => B = 6.84 A bomb is dropped from an aeroplane flying horizontal with a velocity of 720 km/h at an altitude of 980 m. After what time, the bomb will hit the ground? pe [Pe [2=980 Vo VY 98 A horizontal stream of water leaves an opening in the side of a tank. If the opening is h metre above the ground and the stream hits the ground D metre away, then what is the speed of water as it s = 10,/2 sec = 14.14 sec leaves the tank in terms of g, h and D? The given problem is the problem of horizontal projectile. The stream of water shall follow the parabolic path. [2h D t= V9 Ni > [2 ow, ‘vey \2h° SOME APPLICATION OF GENERAL EQUATIONS (A) Projection from a height at an angle 0 above horizontal : u, = u cos @ uy =-—usin @ x =(ucos 0)t a,= +g 1 h = (-u sin 8) t + 3 gt? tie i Eo gt? — (2u sin 6) t- 2h =0 : ee a —— i Fr tet 2usin6 + \(2usine) + 8hg H EA 2g r . wz i fo p_usine , [He] 2h : = + - + (LI °r g9 \\g g a PROJECTILE MOTION Sol. Ex. Sol. (B) Projection from a moving body : Consider a man who throws a ball from a moving trolley. Let the velocity of ball relative to man be u » * » » * * Voalt trolley = Vbat — V trolley Le. Voan = Voall trolley + Viratiey Horizontal component= u cost v Herizontal component = u cos6- v Vertical component= usin 6 Vertical component= u sin 6 Horizontal component = u cos@ Horizontal component = u cos Vertical component =u sing + Vertical component= u sin 4 Y OA=range R M (C) Projection on an inclined plane : a,=-gsinp a,=—g cos B u, =u cos (2-f) u, =u sin (a - B) _ 2usin(a ~B) | ; oO or \-— cose is the time of flight (qurencidian = > *x _ 2u’sin(a —B)cosa 1 x=u,t+ > at? gives the Range |R 2 gcos7B If t, be the time taken by a body to clear the top of a building and t, be the time spent in air, find the ratio t, : t,. t, 2 Total time of flight = 2 (time taken to reach max. height) => t, = 2t, => = ~y 1 When the angle of elevation of a gun are 60° and 30° respectively. The height it shoots are h, and h, respectively. Find the ratio h,/h,. 2 2 aro 2 For angle of elevation of 60°, we have maximum height h, = va 29 - “8g : : : u? sin? 30° u? For angle of elevation of 30°, we have maximum height h, = a Ba g g nh 3 h, 1 410 PROJECTILE MOTION Ex. Sol. Ex. Sol. Sol. From a certain height above the ground, a stone P is dropped gently. Simultaneously, another stone Q is thrown horizontally. Which of the two stones will arrive at the ground earlier? Both the stones will reach the ground at the same time. This result follows from the physical independence of horizontal and vertical motions. A gun, kept on a straight horizontal road, is used to hit a car travelling along the same road away from the gun with a uniform sped of 72 km/hr. The car is at a distance of 500 m from the gun, when the gun is fired at an angle of 45° with the horizontal. Find os le ~ (a) the distance of the car from the gun when the shell . c aS hits it; a Aas Baa — (b) The speed of projection of the shell from the gun. {Ch ree (g = 9.8 m/s*) The speed of the car v = 72 x (5/18) = 20 m/s 2using — uV2 The time of flight of projectile T= “a [as 6 = 45°) «-(1) 2 ai 2 and range of projectile R = u'sin20 Mt wef} g 9g During the flight of shell the car will cover a distance R = 500 + vT Substituting the values of T and R form Eqn. (1) and (2) in the above, 2 [ry 0 OF oH g g or u? — 20,/2u — 4900 =0 = eS la Iza or u = (1/2) [2012 +,/(800 + 4 x 4900] or u=10 [V2+W51] As negative sign of u is physically unacceptable, u = 10 [1.414 + 7.141] = 85.56 mis Substituting the above value of u in Eqn. (2) u? —_ (85.56)* g 98 Two particle located at a point begin to move with velocities 4m/s and 1 m/s horizontally in opposite directions, Determine the time when their velocity vectors become perpendicular. Assume that the motion takes place in a uniform gravitational field of strength g. R= = 746.9 m Velocity of first particle at time t=4i-gtj Velocity of second particle at time t=-—i-gtj Since the dot product of perpendicular vectors is zero therefore (4i-gtj) . (-i-gtj) =0 o|nm or =4 + gt? = 0 or g’t? = 4 or t= 41 PROJECTILE MOTION Ex. Sol. Sol. Particles P and Q of mass 20 gm and 40 gm respectively are simultaneously projected from points A and B on the ground. The initial velocities of P and Q make 45° and 135° angles respectively with the horizontal AB as shown in the fig. Each particle has an initial speed of 49 m/sec. The separation AB is 245 m. Both particles travel in the same vertical plane and undergo a collision. After collision P retraces its path. Determine the position of Q when it hits the ground. How much time after the collision does the particle A B Q take to reach the ground? (Take g = 9.8 m/s?) As the horizontal speed of two particles towards each other is same (u cos45), they will meet at the middle of AB, i.e., at distance (245/2) = 122.5 from A toward B. 2 ei 20; Now as ye a ia g g 9.8 so AB is the range and as the collision takes place at the middle of AB, so it is at the highest point = 245 m of the trajectory. Now applying conservation of linear momentum atthe highest point along horizontal direction keeping in mind, v, = —u, cos45°,. 20 x 10° u cos45° — 40 x 10° u cos45° = — 20 x 10° u cos45° + v,(u = 49 m./sec.) This give v, = 0, i.e., after collision, the velocity of Q at highest point is zero. So Q will fall freely under gravity and will hit the ground in the middle of AB, i.e., 122.5 m from A towards B. _u’sin’?@ — 49%49x1 _ 490 H 6. O_o" TS" = Now as 2g 5gha 3 7 1.25 m 2H 2x 490 5 So time taken by Q to reach ground, t= /=— = es = = 3.536 sec A : | 03) 2 ls it important in the long jump that how much height you take for jumping. It is important in the long jump how high a person jumps. _ u? sin? 0 As h= 2g ut 2h hence g sin? 9 u? sin20 2h ; and range R = ————= x sin 20 = 4h cot 0 g _ssin*@ i.e. the range of jump is determined by initial speed u and angle 0. or height h and angle of projection 0. 12 PROJECTILE MOTION Ex. Sol. Sol. An object of mass 5 kg is projected with a velocity of 20 m/s at an angle of 60° to the horizontal. At the highest point of its path the projectile explodes and breaks up into two fragments of masses 1kg and 4 kg. The fragments separate horizontally after the explosion. The explosion releases internal energy such that the kinetic energy of the system, at the highest point is doubled. Calculate the separation between the two fragments when they reach the ground. (g = 10 m/s*). At the highest point of trajectory applying conservation of linear momentum along horizontal, 5 * 20 cos 60 = 1%*v, +4 v,, Le, v, + 4v, = 50 --(1) And by conservation of energy at highest point, according to given condition, : 1 1 [4 x5x102) x2 = ati +5 4N2, ie. v24+4v3 = 1000 wa (2) Substituting v, from eqn. (1) in (2), (50 - 4v,) + 4v,? = 1000, ie. v,2- 20 v, + 75 =0 which on solving gives v,=5 m/s or 15 mis So from Equn. (1), 30 m/s -10 mis for v, = 5 mis v, , = 15 mis vy, I I and for Vv so if both the particle move in same direction, v,,, = 30-5 = 25 mis and if both move in opposite direction, vy = 15 — (-10) = 25 mis i.e., fragments after explosion separate from each other horizontally 25 m per sec. Now as time taken by fragments to reach, ground _T_usind 20x/3 = = SS oO te a 2 69 4ox2 © So the separation between two fragments when they reach the ground d=v,*xt=25* J3 = 433m If a projectile has a constant initial speed and angle of projection, find the relation between the change in the horizontal range due to change in acceleration due to gravity. ; u? sin 20 Horizontal range, R = —g Differentiating t w.r.t. we have m. _w sin 20 . dg o [-. u and @ are constant] u’ sin 26 dg dg dR=-—z—_— =-R_ = 9g 9g g aR __0g or R 7 13