Q PROJECTILE MOTION 1. BASIC CONCEPT : 14 12 13 Projectile ‘Any object that is given an initial velocity obliquely, and that subsequently follows a path determined by the net constant force, (In this chapter constant force is gravitational force) acting on itis called aprojectile Examples of projectile motion © Accticket ball hit by the batsman for a six © Abullet fired from a gun. ‘© Apacket dropped from a plane; but the motion ofthe aeroplane itself isnot projectile motion because there are forces other than gravity acting on it due tothe thrust of its engine. Assumptions of Projectile Motion ‘© Weshall consider only trajectories that are of sufficiently short range so thatthe gravitational {force can be considered constant in both magnitude and direction. © Alleffectsof air resistance will be ignored. © Earth is assumed to be flat Projectile Motion : © The motion of projectile is known as projectile motion © Itisanexample of two dimensional motion with constant acceleration. © Projectile motion is considered as combination of two simultaneous motions in mutually perpendicular directions which are completely independent from each other i.e. horizontal motion and vertical motion. aso cosa Horizontal using motion Vertical motion Parabolic path = vertical motion + horizontal motion. Galileo's Statement : ‘Two perpendicular directions of motion are independent from each other. In other words any vector quantity directed along a direction remains unaffected by a vector perpendicular t it.PROJECTILE THROWN AT AN ANGLE WITH HORIZONTAL 2.4 | ocoaey oe aT Consider a projectile thrown with a velocity u making an angle # with the horizontal Initial velocity uis resolved in components in a coordinate system in which horizontal direction is. taken as x-axis, vertical direction as y-axis and point of projection as origin u,=ucosé u,=usin ‘Again this projectile motion can be considered as the combination of horizontal and vertical mation. Therefore, Horizontal direction Vertical direction (2) Initial velocity u,=ucos 0 Initial velocity u, = usin (©) Acceleration a, =0 Acceleration a, =g (¢) Velocity after time t, v,= u cos Velocity after time tv, = usin @—gt Time of flight = The displacement along vertical direction is zero for the complete flight, Hence, along vertical direction net displacement = 0 2usiné > > a Horizontal range 2usino Rout = Reucose. > Maximum height : {At the highest point ofits trajectory, particle moves horizontally, and hence vertical component of velocity is zer0. Using 3" equation of mation i.e. v= uP +2as we have for vertical direction uesinto-2gH > Resultant velocity : cos 6 j + (usin @gt) j zl Where, | = yutcos®e + (usine — gt)? and tana=v,!v, Also,Note : © Results of article 2.2, and 2.3 are valid only if projectile lands at same horizontal level from Which it was projected. © Vertical component of velocity is positive when particle is maving up and vertical component of velocity is negative when particle is coming down if vertical upwards direction is taken as positive, 2.5 General result : © Formaximumrange @= 45° Rup © We get the same range for two angle of projections u and (90 ~ a) but in both cases, maximum heights attained by the particles are different u? sin26 This is because, R and sin 2 (90 —a) @ eof Re u? sin26 vu? sin® t ie. = ot ‘© Range can also be expressed as _utsing@_ gusinoucoso _ 2st 9 9 9 Example 1. A body is projected with a speed of 30 ms~' at an angle of 302 with the vertical. Find the maximum height, time of flight and the horizontal range of the motion. [Take g = 10 mis*] Solution: Here u= 30 ms", Angle of projection, 0 = 90-30 = 60? Maximum height, uesin?a 30%sin®60" 900 3 _ 135 29 3x10 204" 4 ™ 2usind _ 2x30xsin60° Time of fight 7 = ° 30x30 x 2sin 60° cos 60" ia Horconalvange= ix W802, SxBOr2ANENEOEED 46 my Example 2. A projectiles thrown with a speed of 100 m/s making an angle of 60° with the horizontal. Find the time after which its inclination with the horizontal is 45° ? Solution : u,= 100 x cos60° = 50 100 x sin60° = 503 When angles 45", v tan 45° =Example 3. Solution : Example 4. Solution : Example 5. Solution : large number of bullets are fired in all directions with the same speed v. What is the maximum area, ‘on the ground on which these bullets will spread ? Maximum distance up to which a bullet can be firedis its maximum range, therefore The velocity of projection of a projectile is given by : i = 5i+ 10]. Find (a) Time of fight, (b) Maximum height, (c) Range Wehaveu, u,=10 2using (a) Time of flight = vesin?o uy? 10x10 ©) Maximum height = SO = Bo= Soy = 5m 2usind.ucos® (Range = Sn + om A particle is projected at an angle of 30° wrt. horizontal with speed 20 mis () Find the position vector of the particle after ts. (i) Find the angle between velocity vector and position vector at () x =ucosat 3 220 Bint = 10/5m 1 yeusinat- > «10% z = 20% bx (5 (ny =5% =20% 4 <(1)-8 1") =m Position vector, F = 10N81+8), 17 yhovaP +5? “ WF =1VIIF [eos uF 300 > 0089 = Tey = Toa vees = ences («(5)Q 3. EQUATION OF TRAJECTORY ‘The path followed by a particle (here projectile) during its motion is called its Trajectory. Equation of trajectory isthe relation botwoon instantanoous coordinates (Hore x & y coordinate) ofthe particle if we consider the horizontal tection, t ‘cos 8.1 “) Forvertical direction y 1, t- V2 gt Using. t— 12g ) Eliminating from equation (1) & (2) = xtano- 9% wees This s an equation of parabola called as trajectory equation of projectile mation. Other forms of trajectory equation 9x2(1+tan?0) © yextano~ SY gx? © yexteno- Frc0a7e 1-_ = xtan@ ‘2u? cos* btand. tang |'-——*—_ > an |! BF ina oo x = yextano [1-2] —— Solued Examples Example 6. Find the value of @ inthe ciagram given below so that the projectile can hit the target. = 10m _ 9x2(1+tan? 9) Solution y= xtan@. a A 5x(20 210 20tan8— “Foz (1+ tan) 2 2a dtanb—(t stan) > tan? 4 tan B43 0 = 0-45", tan-(3) = (tan 6-3) (tana—1Example 7. ‘A boall is thrown from ground level so as to just clear a wall 4 m high at a distance of 4 m and falls at aistance of 14 m from the wall. Find the magnitude and direction of initial velocity ofthe ball Figure is given below. Solution p in om > The ball passes through the point P(4, 4) Also range = 4+ 14= 18 m. The trajectory ofthe the ballis, x tano(1-2 ) 1-4 Zz Now x=4m,y=4mandR = 18m 4edtano Ig ]=a4tana. 5 o tanda2 And R= BESO 7 9 or > 182 o 4, PROJECTILE THROWN PARALLEL TO THE HORIZONTAL FROM SOME HEIGHT Consider a projectile thrown from point O at some heighthtrom the ground og with a velocity u. Now we shall study the characteristics of projectile motion by resolving the mation along horizontal and vertical directions. QO i) 44 4.2 4.3 Horizontal direction Vertical direction Initial velocity u, Initial velocity u, =0 Acceleration a, =0 Acceleration, =g (downward) Time of flight = ‘This is equal tothe time taken by the projectile to return to ground. From equation of motion 1 Uut+ 5 at? , along vertical direction, we get -h de ge 1 oe s+ Teae > he Sat > t Horizontal range Diane coved ths ete lng hora recon bent pont pjetont thopoton tho A au cut a Velocity at a general point P(x, y) : velocity of projectile in vertical direction aftr timet v= 0+ (o)t=-at=gt (downward) v= yu? +0"? and tana44 45 Solved Examples Velocity with which the projectile hits the ground : V,=u Viz= 08—2g(-h) V,= 20h ve ow = Veda gn Trajectory equation The path traced by projectile is called the trajectory. Attertime', ut «y Fae ) From equation (1) t= xia Putthe value of tn equation (2) Tag. a yg This is trajectory equation of the particle projected horizontally from some height. Examples based on horizontal projection from some height : Example 8. Solution : Example 9. A projectiles tired horizontally with a speed of 98 ms“ fram the top of ahill 490 m high. Find ()) the time taken to reach the ground (i) the distance of the target from the hill and (i the velocity with Which the projectile hits the ground. (take g = 9.8 m/s*) (The projectiles fred from the top O of a hill with speed u = 98 ms" along the horizontal as shown as OX. It reaches the target P at vertical depth (inthe coordinate system as shown, OA =y = 490 m As, y us 0m 490-4 a8 wt ot 400 . @ Distance ofthe target rom the his givenby, ’ AP = x= Horizontal velocity time = 98 x 10 = 980 m (ii) ‘The horizontal and vertical components of velocity v of the projectile at point P are v,=U=98ms" V) =U, +gt=0+9.8 x 10=98 ms" Ve har = (98? 98 98/2 ms" Nowitthe esuitant velocity v makes an angle with the horizontal then yy 98 v, 98 tanp 458 ‘A motorcycle stunt rider rides off the edge of a cliff. Just at the edge his velocity is horizontal, with ‘magnitude 9.0 m/s. Find the motorcycle's position, distance from the edge of the cliff and velocity, after0.5s.Solution : Att =0.50 s, the x and y-coordinates are roa) v,t= (8.0 mis) (0.50) 4 et . 7 Fat == 5 (10 mist (0.50 5} ‘The negative value of y shows that this ime the motorcycle is below its starting point. ‘The motorcycle’s distance from the origin a this time. The components of velocity at this time are Yovy= 80mis 10 mis?) (0.0 s) = -5 mis. The speed magnitude of he velocity) atts mes = (Ene = [eomia? comin? = sA06 mis Example 10. _An objects thrown between vo tal buildings. 180m rom each othe. The obectisthrown horizon: tally om awincow 88 m above gound om onebuing tough a window 10.9 m above glound hy the other building. Find out the speed of projection. (use g = 9.8 m/s*) sotion: [PR (oeaaa olution : a us 60 mis aQ 6. PROJECTION FROM A MOVING PLATFORM CASE(1): sol andor bor When a ball is thrown upward from a truck moving with uniform speed, then observer A standing in the truck, will see the ball moving in straight vertical line (upward & downward) The observer B sitting on road, will see the ball moving in a parabolic path. The horizontal speed of the ball is equal to the speed of the truck.CASE (2): CASE (4): CASE(5): When a ball is thrown at some angle ‘8" in the direction of motion of the truck, horizontal & vertical component of balls velocity wart. observer A standing on the truck, is ucos®, and usiné respectively Horizontal & vertical component of bal’s velocity wrt. observer B sitting on the ground, is U,=Ucos® + v and u,=usind respectively. . rode Birr When a ball is thrown at some angle ‘0 inthe opposite direction of motion of the truck, horizontal & vertical component of ball’s velocity wrt. observer A standing on the truck, is ucos®, and usin® respectively. Horizontal & vertical component of bal’s velocity wrt. observer B sitting on the ground, is U,=Ucos® —v and u,=usin® respectively. usin x, hie be When a ballis thrown at some angle ‘from a platform moving with speed v upwards, horizontal & vertical component of ball’s velocity wrt. observer Astanding on the moving platform, is ucos8 and Usin@ respectively. Horizontal & vertical component of bal’s velocity wrt. observer B sitting on the ground, is u,=ucos® and u, =usind + v respectively. Vor When aballis thrown at some angle’ from a platform moving with speed v downwards, horizontal & vertical component of balls velocity wrt. observer A standing on the moving platform, is ucos® and usind respectively. Horizontal & vertical component of ball’s velocity w.t. observer B siting on the ground, is ,=ucos8 and u,=usin®—v respectively cose d, ir—— Solued Examples Example 11. Solution : Example 12. Solution : ‘A boy standing on a long railroad car throws a ball straight upwards. The car is moving on the horizontal road with an acceleration of 1 m/s* and the projection speed inthe vertical directions 9.8 mis. How far behind the boy will the ball fall on the car ? Let the intial velocity of car be u time of fight, 2u, t-—ta2 9 \where _u, = component of velocity in vertical direction Distance travelled by car 4 2 unde d xtae 2u+2 distance travelled by ball x sux2 A fighter plane moving with a speed of 50/2 m/s upward at an angle of 45% with the vertical, releases a bomb. Find (a) Time of ight (b) Maximum height of the bomb above ground 1 sont? 1000 = s0t— 4«10%t* 1900 = Sot 5 10xt 108 - 20 (t=20) (t+ 10)=0 0 sec 50% 50x wae 50x50 2%" 23025" Hence maximum height above ground H= 1000 +125 = 1125mA particle moves along the parabolic path y = ax? in such a way that the y-component of the velocity remains constant, say ¢. The x and y coordinates are in meters. Then acceleration of the particle at x=tmis © -z (A) ack (8) 2ac?j particle is projected with a velocity 'u' in horizontal direction as shown in the figure. Find u' so that the particle collides orthogonally with the inclined plane of the fixed wedge. (20h a [aan V2rc0s% (8) Var sinte © freer 0) Virsin® particle is projected from the horizontal x-z plane, in vertical x- plane where x-axis is horizontal and positive y-axis vertically upwards. The graph of 'y' coordinate of the particle v/s time is as shown. The range of the particle is 3 m. Then the speed of the projected particle is (a) WB mis © \ems (©) v5 mis (0) 28 mis Velocity of a stone projected, 2 second before it reaches the maximum height, makes angle 53° with the horizontal then the velocity at highest point willbe (A) 20 mis (6) 15 ms (©)25 mis (0) 8073 mis ‘Apebble is thrown horizontally from the top of a20 m high tower with an initial velocity of 10 mis. The air drag is negligible. The speed of the pebble when its at the same distance from top as well as base of the tower (g= 10 mist) (A) 10y2 mis (8) 10¥3 mis (©)20ms (0)25 ms projectile Ais projected from ground. An observer 8 running on ground with uniform velocity of magnitude 'v observes Ato move along a straight ine. The time of fight of Aas measured by B is T. Then the range R of projectile on groundis, ()R=vT (8)RevT — ()RovT (0) information insutticient to draw inference A all is projected with velocity u at right angle to the slope which is inclined at an angle a with the horizon: tal. The distance x’ along the inclined plane that it will travel before again striking the slope is - ny 2a cose tana 2u? tana au? tana. ws OF (9 cosa ©)"F sina10. 1. 12. 13. 14, 18. ‘Two guns are mounted (fixed) on two vertical cliffs that are very high from the ground as shown in figure. ‘The muzzle velocity of the shell from G, is U, and that from G, is u,. The guns aim exactly towards each other The ratio u, : u, such that the shells collide with each other in air is (Assume that there is, no resistance of air) “a (ayn :2 @1:4 (C) will not collide for any ratio (0) will collide for any ratio. ‘Aball is thrown upward at an angle of 30° with the horizontal and lands on the top edge of a building that is 20 m away. The top edge is 5m above the throwing point. The inital speed of the ball in metrol second is (take g = 10 m/s?) (A) 10 mis (8) 20 mis (©) 25 mis (0) 30 mis A ball is thrown from bottom of an incline plane at an angle « from the inclined surface up the plane. Another ball is thrown from a point on the inclined plane with same speed and at same angle a from tho inclined surface down the plane. If in the two cases, maximum height attained by the balls with respect to the inclined surtace during projectile motion are h, and h, then (A)hy > he ()h, is 2 Re z Oy oy Ow “Two porsons X and ¥ are playing with wo diferent balls of masses m and 2m. IX throws hisball vertically up and atan angle 0 with vertical both of them stay in airfor the same time. The maximum heights attained by tho two are inthe ratio (aast (B) 1: cos® or (0) 1 5008 Agunis pointed towards a 100 m high target as shown in figure. Target is released at the same time when the gun is shot. To hit the target in air, muzzle speed should not less than (g = 9.8 mis*) Target | 100m aun, Son (72 mis (8) 1445 mis (©) 10m (0) 1070 mis Atruitis fling with ero inital velocity trom a 50 m high tree. Aman, which sat 0 m horizontal distance from the tee fire abulletwith a speedo 49/2 misec. at an angle 45° wth horizontal After how much time will tne bullet hit the fruit 2 Treo (A) 2 sec, (8) 5 sec, (C)4 sec, (0) None of these Arlicoptor's flying at an altitude of 200 m with 25 m/s at an angle of 37° above horizontal when a package is dropped from it. After how much time (in sec) package hits the ground : (g = 10 mist) a7 (68 9 (0)10KVPY PROBLEMS (PREVIOUS YEARS) A boy standing on the foothpath tosses a ball straight up and catch it. The driver of a car passing by moving with uniform velocity sees this, IKVPY_2009] The trajectory ofthe ball as seen by the driver will be, a4 ®) f) © r” According to the quantum theory, a photon of electromagnetic radiation of frequency v has energy E = hv where his known as planck’s constant. According to the theory of relativity, aparticle of mass mhas equivalent energy E = me?, where c is speed of light. Thus a photon can be treated as a particle having hv effective mass m= >. fa flash of lightis sent horizonatally in earth’s gravitational field, then photons while © traveling a horizontal cistance d would fal trough a distance given by IKVPY_2009 2 Marks} ot a med? , oz On om (0) zero A firecracker is thrown with velocity of 30 ms” in a direction which makes an angle of 75* with the vertical axis. At some point on its trajectory, the firecracker split into two identical pieces in such a way that one piece falls 27 m far from the shooting point. Assuming that all trajectories are contained in the same plane, how far will the other piece fall from the shooting point ? (Take g = 10 msand neglect air resistance) IkvPY_2012] (A)63mor144m — (B)28mor72m (©)72mor99m (0)63mor 117m 18. EXERCISE-1 Oo 2 #2 © 4 ® & ® &6& A 2 O Oo. ®» 20 On, O 2 O B O MW A oOo 6 A 7 © 8B 6 1% O wm A EXERCISE-2 oO 2 # 3. oO1 2 0=u.coso 9 sinot {along the plane equation => uegttana 20h V" z V2sc05% 3. Fromgraph us tan 60° = /8 mis anos wet 2xuexV ange Re EE oy p= Pe oru,=5mis = id sud = V8 mis 4. Twosecond before maximum height, «932 =20 nis 20 e20 sang = at maximum height v= v, = 15 m/s “ ‘The pebble is at same distance from top and base of the tower after it falls down by a distance h = 10m. Vin Ute 2gh= 100425101 0 V3 mis.6 The horizontal and vertical components of initial velocity of projectile are as shown in figure. Since the observer moving with uniform velocity v sees the projectile moving in straight line ‘ 8 i vay ofan gon ‘elo other tame as seat ra “Thotine of ight as measuredty oserverBis uning 3 enc aio angoat rjcta an rund R=(ucose)T =vT te 1. Emtontormetonpopendcuaro thence = ut-# pease. t= hy = hy1. 12. 13. 4 18. 16. trajectories are identical 3.5 mist (Easy) Itis obvious from the figure shown thatthe angle between the velocity vectors 1 sec before and 1 sec after it attains maximum height is 90° toms tetsec after te1s00 before, 10rms Leta (ucisu,) misbe veloc of preecton Atheighth=0.4m, ¥=(6i+2i) ms As horizontal component of velocity remains constant u,=6mis ‘Taking motion in y-direction, vi ut -20h = uF -24 10x04 By M21 hone, tan = i=“ F-= Te 0-30" (Easy) Time of fight and maximum height depends on vertical component of intial velocity. Time of tight is less in case | so it hits target first. Let v be the speed at a height h above A, the speed at height h below Ais 2v. I uis initial speed at A then ve=u?2gh and (2v)*=u" + 2gh 10 solving ut= gh ay If His max height above A, then ut 2gH @ from 1 and2 Both reach the highest point same time, hence thei inal vical component must be same. 008308» v, “2 ve V3"7. 18. 19. ForX. :T, Again, for X: H, usin? (90-8) 29 Ht He ~ woos™6 In the limiting condition gun wil hit the target just before it its the ground uP sin2x45¢ = Range = 100m= lange = 100 m 3 > us 00x98 = 1405 mis It we observe relative tothe fruit, Time to strike the bullet, But actual time of flight ofthe bullets ex 4 using 2*1002 9 10 soe So, after 2 sec., the bullet will hit the ground, so it cannot hit the fruit = 200 = 25 sins7°t— 210) 8 sec.