Motion in a Plane Projectile Motion

Motion in a Plane PROJECTILE MOTION MOTION IN TWO DIMENSIONS (i) In a two dimensional motion, the motion takes place in a plane. (ii) If the plane in which motion takes place is taken as the (X-Y) plane, the coordinates of a point in this plane will be in this plane will be represented by (x, y) and the position vector r will be  r  x ˆi  y ˆj (vii) Motion with uniform velocity : Suppose the coordinates of a particle at t = 0 are (x0, y0) and after time t the coordinates become (x, y), then for uniform velocity      r  r0 r  r0  v t0 t    or r  r0  v t  x î  y ĵ  ( x 0 î  y 0 ĵ)  ( v x î  v y ĵ) t  ( x 0  v x t ) î  ( y 0  v y t ) ĵ or x = x0 + vxt and y = y0 + vy t (viii) Motion with constant acceleration :  dv  (iii) If (x1, y1) are the coordinates of the position of a particle at time t1 and (x2, y2) are the coordinates of the particle at time t2 then  r1  x1ˆi  y1ˆj and  r2  x 2 ˆi  y 2 ˆj     r  r2  r1   If the velocity is v 1 at time t1 and it becomes v 2   at time t2, then dv  a dt  v2 Integrating The displacement of the particle will be  d2 r   (a) If a is constant acceleration then a  dt dt 2   v1 t  2 dv  a dt  t1    or v 2  v 1  a( t 2  t 1 )    v 2  v1  a t 2  t1 = (x2 – x1) ˆi  (y 2  y1 ) ˆj (b) Let the velocity at time t = 0 be u and the velocity at time t be v .      v u v u   a t0 t    v  u  at or The components of v will be (iv) The magnitude of displacement is |  r | = [(x2 – x1)2 + (y2 – y1)2]1/2  (v) At time t if the position is given by r  xiˆ  yjˆ , vx= ux + axt and vy = uy + ayt the instantaneous velocity of the particle at this position will be Integrating from initial position r  r0 at t   dr  dx  ˆ  dy  ˆ v  i j dt  dt   dt  = v x ˆi  v y ˆj (c) v  ( v 2x  v 2y )1/ 2 (vi) If the particle is displaced from position  r1 at  time t1 to the position r2 at time t2 then average  velocity during the interval (t2 – t1) is v av   r r  2 1 t 2  t1    dr  (u  at)dt    = 0 to position r at time t  r Speed or magnitude of velocity will be   dr   v  u  at dt  t    dr   (u  at)  r0 dt 0      1 2  r  r0  u t  a t 2 In terms of Cartesian components x = x0 + uxt + 1/2 ax t2 y = y0 + uyt + 1/2 ay t2  1 2 or r  r0  u t  a t 2 PROJECTILE MOTION Ground to ground projection : Consider the motion of a bullet which is fired from a gun so that its initial velocity u makes an angle  with the horizontal. Take x-axis along ground and y-axis along vertical.  u x = u cos  (along u can be resolved as horizontal) & uy = u sin  (along vertical) motion of bullet can be resolved into horizontal and vertical motion. * In horizontal direction there is no acceleration so it moves with constant velocity vx = ux = u cos  So distance traversed in time t is x = ux t or x = (u cos ) t or t  x u cos  . Motion in a Plane * The trajectory of projectile is parabolic The projectile will rise to maximum height H (where vx = u cos , vy = 0) and the move down again to reach the ground at a distance R from origin. Setting x = R and y = 0 (since projectile reaches ground again) O = R tan  – g 2 2 2u cos  .R 2 2 2 We get R  2u cos   sin  g cos  2 or R  2u . sin  cos  or Range g R u2 sin2 g ....(i) * If time for upward journey is t at highest point vy = 0 so 0 = (u sin ) – gt (vy = uy – gt) or t  u sin  g  T = 2t (it will take same time for downward journey) * The motion in the vertical direction is the same as that of a ball thrown upward with an initial velocity uy = u sin  and acceleration = – g downward. So at time t vertical component of velocity vy = uy – gt = u sin  – gt ..........(ii) Displacement along y direction y = (u sin ) t – 1/2 gt2 ..........(iii) Substituting the value of t from eq. (i) in eq. (iii) we get y = (u sin or y = x tan  – parabola.  x  1  x    g  )   u cos   2  u cos   g 2 2 2u cos  2 . x 2 . This is eq. of  Time of flight T  2usin  g * At the highest point y = H and vy = 0 So that H  u 2y 2g v 2y  u 2y  2gy or Maximum Height H u2 sin2  2g * we can also determine R as follows x = ux t so R = ux . T  Motion in a Plane (3)If angle of projection is changed from  2u sin   = (u cos )  g   2 or R  u sin 2 g velocity at time t to   '  (90  ),  v t  v x t ˆi  v y t ˆj  (ucos )iˆ  (u sin   gt)ˆj 2 2 v  u cos   (u sin   gt ) then range u2sin2(90) u2sin2'   u2sin2   R R'      g g g 2 Note : (i) Alternative equation of trajectory y = x tan x   1    R 2 where R  2u sin  cos  g (ii) Vertical compound of velocity vy = 0, when particle is at the highest point of trajectory. So a projectile has same range for angles of projection  and (90 – ) (iii) Linear momentum at highest point = mu cos  is in horizontal direction. But has different time of flight (T), Maximum height (H) & trajectories (iv) Vertical component of velocity is + ive when particle is moving up. Range is also same for 1 = 45º –  and 2 = 45º (v) Vertical component of velocity is –ive when particle is moving down. (vi) Resultant velocity of particle at time t v  v 2x  v 2y at an angle  =  vy tan–1   vx  .   (vii) Displacement from origin, s  x 2  y 2 + .  u 2 cos 2  equal  g   (4)For maximum range u2 R = Rmax  2 = 90º for  = 45º Rmax = g [For sin 2 = 1 = sin 90º or  = 45º] When range is maximum Then maximum height reached Special Points : H (1) For projectile motion give : T 2u sin  g R u2 sin 2 g H u 2 sin2  2g (2)In case of projectile motion The horizontal component of velocity (u cos ), accleration (g) and mechanical energy remains constant. Speed, velocity, vertical component of velocity (u sin ), 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. u2 sin2 45º (WhenRmax ) 2g or H  u2 4g Hence maximum height teached (for Rmax) H  Rmax 4 (5)For height H to be maximum H u 2 sin2  = max 2g i.e. sin2 = 1 (max)or for  = 90º u2 Hmax = 2g When projected vertically (i.e. at  = 90º) in this case So that Range R u2 sin(2  90º ) u2 sin180º  0 g g u 2 For vertical projection Hmax = 2g and For u2 oblique projection with same velocity Rmax = g so Hmax = R max 2 If a person can throw a projectile to a maximum distance (with  = 45º) Rmax u2 = g Motion in a Plane HORIZONTAL PROJECTILE MOTION TOP TO GROUND PROJECTION 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 origin O is at point of projection as shown is fig. The maximum height to which he can throw the projectile (with  = 90º) Hmax = R max 2 (6)At highest point Potential energy will be max and equal to (PE)H = mgH = mg . u2 sin2  2g or (PE)H = 1 2 mu2 sin2 . While K.E. will be minimum (but not zero) and as at the highest point the vertical component of velocity is zero. (KE)H = 1 2 mv H2 = so (PE)H + (KE)H = = 1 2 1 m(u cos )2 = 2 1 mu2 sin2  + 2 1 2 1 2 mu2 cos2  mu2 cos2  mu2 = Total Mechanical energy So in projectile motion mechanical energy is conserved. 1 mu 2 sin 2   PE  2  tan 2     1 KE 2 2  H mu cos  2 So if  = 45º     tan2  PE = K.E. = 1 2 =1 M.E. at highest point i.e. if a body is projection at an angle  = 45º 10 the horizontal then at highest point, half of its M.E. is K.E. and half is P.E. (7)In case of projectile motion if range R is n times the maximum height H, i.e. R = nH u2 sin2  u2 sin 2  n. then 2g g or tan  = 4 n  or 2cos  =  n. sin  2 4 = tan–1  n    (8)Weight of a body in projectile motion is zero as it is a freely falling body. 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 i.e velocity of projection and without acceleration. The horizontal component of velocity vx = u and horizontal displacement x = u . t .....(i) displacement along vertical direction is y to calculate y, consider vertical motion of the projectile initial velocity in vertical direction uy = 0. acceleration along y direction ay = g (acceleration due to gravity) So vy = ay t (y component of velocity at time t) or vy = gt .....(ii) (as body were dropped from a height)  Resultant velocity at time t is v  v î  (gt ) ĵ v  u2  (gt)2 if  is the angle of velocity with x-axis (horizontal) tan   1 2 gt u and y  gt 2 ....(iii) 1 x   or y  2 g.  u    or y  g 2u here k  2 . x2 g 2u 2 2 [from equation (i) t = or y = kx2 (k is constant) This is equation of a parabola. x u ] Motion in a Plane A body thrown horizontally from a certain height above the ground follows a parabolic trajectory till Ex.1 When the angle of elevation of a gun are 60º and it hits the ground. 30º respectively. The height it shoots are h1 and h2 (i) Time of flight T  2H [as y = 1/2gt2 T  2H ] respectively. Then h1/h2 equals to. g g (ii) Range  horizontal distance covered = R. (1) 3/1 (2) 1/3 R = u × time of flight (3) 1/2 (4) 2/1 R  u. 2H g [ H = g 2u 2 Sol. For angle of elevation of 60º, we have maximum R2] (iii)Velocity when it hits the ground v g  u 2  2gH PROJECTILE MOTION ON AN INCLINED PLANE u2 sin2 60 º 3u 2  2g 8g height h1  For angle of elevation of 30º, we have maximum height h 2  h1 3  h2 1 u2 sin2 30º u2  2g 8g Hence correct answer is (1) (can only be solved after understanding conservation of linear momentum.) Time of Flight Vertical displacements on inclined plane = 0 Sy = Uyt + 1 2 ay t 2 0 = Usin ( – ) t –1/2 (g cos )t2 t 2U sin(    ) g cos  t check by  = 0 2U sin  g Which is time of flight on horizontal plane. Range on inclined plane Sn = Uxt + 1/2 ax t2 (O O’) = U cos ( – )t – 1/2 g sin  t2 substituting the value of (t) = U cos (  – ) ×  2U sin(   )  2U sin(   ) 1  g sin  .   g cos  2  g cos   (OO’) = R = 2 How much time after the collision does the particle Q take to reach the ground.? {Take g = 9.8 m/s2) 2U2 sin(    )(cos ) Sol. As the horizontal speed of two particles towards each other is same (u cos 45), they will meet at the middle of AB, i.e., at distance (245/2) = 122.5 from A toward B. g cos 2  condition for maximum range  2 sin(    )(cos )   g cos 2   R = U2 .  = U2  sin(2   )  sin     for R sin (2 – ) = 1 max g cos 2    2  –  = /2 2 Rmax = U [1  sin2 ] =  g[1  sin  ]  2 Ex.2 Particle P and Q or 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 45º and 135º angles respectively with the horizontal AB. Each particle has an initial speed of 49 . 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. = /2 + /4 2 U U Rmax= g ( 1 sin  )  g(1  sin ) Now as R u2 sin 2 u2 sin 2 49  49  1 R   245 m 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 of the trajectory. Now applying conservation of linear momentum at the highest point along horizontal direction keeping in mind, vp = – up cos 45º 20 × 10 –3 u cos 45º – 40 × 10 –3 u cos 45º = – 20 × 10–3 u cos 45º + vQ This give vQ = 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. Now as u2 sin2  49  49  1 490 H    61.25m 2g 2  9.8  2 8 Motion in a Plane According to given problem, R = 500 + vT Substituting the values of T and R from eq. (1) and (2) in the above, u 2 u2  20 = 500 + g g or u2 – 20 2 u – 4900 = 0 or u = (1/2) [20 2 ± (800  4  4900 ) ] or u = 10[ 2 ± 51 ] As negative sign of u is physically unacceptable, u = 10 [1.414 + 7.141] = 85.56 m/s Substituting the above value of u in eq. (2) R So time takes by Q to reach ground, u2 (85.56) 2   746.9 m g 9.8 Ex.4 A person observes a bird on a tree 39.6 meter high and at a distance of 59.2 meter. With what t velocity the person should throw an arrow at an angle (can only be solved after understanding conservation of 45º so that it may hit the bird ? of linear momentum.) Sol. Let u m/s be the initial velocity of the arrow and 2H 5  2  490  s  3.536 s    g 2  8  9 .8  Ex.3 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 speed 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 45º with the horizontal. Find (a) the distance of the car from the gun when the shell hits it. let it hit the bird after t second. The horizontal component of velocity u is u cos 45º = u/ 2 . Hence the horizontal distance travelled in t second u is 2  t  59.2 The vertical distance travelled in t second is 39.6 m, acceleration g = – 9.8 m/s2 and initial vertical velocity = u sin 45º = u/ 2 . Hence by the formula (b) The speed of projection of the shell from the gun. (g = 9.8 m/s2) h = ut + Sol. The speed of the car v = 72 × (5/18) = 20 m/s we have The time of flight of projectile T  2u sin  u 2  g g [as  = 45º] ......(1) and range of projection R u2 sin 2 u2  ....(2) g g ........(i) 1 2 gt2, 39.6  u 2 t 1 (9.8)t 2 2 .......(ii) by eq. (i) and (ii), we get 39.6 = 59.2 – 1/2(9.8) t2 o 2 or t  59.2  39.6 19.6  4 4.9 4.9 t = 2 seconds  Substituting this value of t in eq. (i), we get u 2   2  59 .2 u = 29.6 2 m/s Ans. Ex.5 An enemy aircraft is flying at a height of 875 m from the surface of earth with a uniform velocity of 150 m/s in horizontal direction. When this aircraft is just over an antiaircraft gun, a shell is fired from the antiaircraft gun with velocity u at an angle  above the horizontal direction. This shell hits the aircraft 35 second after being fired. Determine u and  . g = 10 m/s2 and tan 53º = 4/3. Sol. The shell will hit the aircraft provided the horizontal velocity of the shell u cos, is equal to the horizontal velocity of the aircraft. That is u cos  = 150 ms–1 .....(i) The vertical distance covered by the shell is h = 875 m, acceleration is g = – 10 ms–2 and the initial vertical velocity of the shell is u sin . The shell hits the aircraft in t = 35 s. Putting these values in the relation h = ut + 1/2 gt2, we have 875 = (u sin ) × 35 + 1/2 (–10) × (35)2 Solving : u sin  = 200 ms–1 ....(ii) From eq. (i) and (ii) u2 = (150)2 + (200)2 = 22,500 + 40,000 = 62,500.     u = 250 ms–1 Dividing eq. (ii) by eq. (i), we get tan     200 4  150 3 = tan–1 (4/3) = 53º Ex.6 A bomb is dropped on an enemy post by an aeroplane flying with a horizontal velocity of 60 km/ hr and at a height of 490 m. How far the aeroplane must be from the enemy post at the time of dropping the bomb, so that it may directly hit the target ? (g = 9.8 m/s2) What is the trajectory of the bomb as seen by an observer on the earth ? What as seen by a person sitting inside the aeroplane ? Sol. The horizontal and vertical velocities of the bomb are independent to each other. The time taken by the bomb to hit the target can be calculated by its vertical motion. Let this time be t. Putting h = 490 m and g = 9.8 m/s2 in the formula h = 1/2 gt2, we have 490 = 1/2 × 9.8 × t2  t 2  490  10 s 9.8 Motion in a Plane The bomb will hit the target after 10 second of its dropping, The horizontals velocity of the bomb is 60 km/h which is constant. Hence the horizontal distance travelled by the bomb in 10 second (horizontal velocity × time) is 60 km/h × 10 s = 60 km/h × 10 60  60 h = 1/6 km. Hence the distance of aeroplane from the enemy post is 1/6 km = 1000/6 = 500/3 meter. The trajectory of the bomb as seen by an observer on the ground is parabola. Since the horizontal velocity of the bomb is the same as that of the aeroplane, then falling bomb will always remain below the aeroplane. Hence the person sitting inside the aeroplane will observe the bomb falling vertically downwards. Ex.7 A bomb is fired from a cannon with a velocity of 1000 m/s making an angle of 30º with the horizontal (g = 9.8 m/s2). (i) What is the time taken by the bomb to reach the highest point ? (ii) What is the total time of its motion ? (iii) With what speed the bomb will hit the ground and what will be its direction of motion while hitting ? (iv) What is the maximum height attained by the bomb ? (v) At what distance from the cannon the bomb will hit the ground ? Sol. (i) Let u be the initial velocity of the bomb and 0 its angle of projection. The time taken by the bomb to reach the highest point is given by u sin  0 1000  sin 30 º 1000  1/ 2    51sec . g 9.8 9.8 t (ii) The total time of its motion is, T = 2t = 2 × 51 = 102 seconds. (iii) The bomb will hit the ground with the same speed with which it was fired. Hence its speed of hitting = 100 m/s. Also, the angle of hitting with respect to the horizontal is 30º (iv) The maximum height attained by the bomb is h º  u 2 sin2 0 (1000)2  sin2 30º  2g 2  9. 8 (1000)2  (0.5)2  1.27  10 4 meter 2  9.8 (v) Horizontal range is R u2 sin 20 (1000)2  sin 60º   8.83  104 meter g 9.8 Ex.8 A ball is thrown from the top of a tower with an Motion in a Plane horizontal component, ux = u cos 0 = 10 cos 30º initial velocity of 10 m/s at an angle of 30º above the = 8.65 m/s and vertical component (upwards), horizontal. It hits the ground at a distance of 17.3 m uy = u sin 30º = 5.0 m/s from the base of tower. Calculate the height of the If the ball hits the ground after t seconds of projection, tower. (g = 10 m/s2) Sol. then the horizontal range is R = ux × t = 8.65 t meter  t R 17 .3m   2 .0 s 8.65 8.65m / s If h be the height of the tower, then h = u’y t + 1/2 gt2, where uy’ is the vertical component (downward) of the velocity of the ball The angle of projection of the ball is 0 (= 30º) and the velocity of projection is u (= 10 m/s). Resolving u in horizontal and vertical components, we have Here uy ‘ = – uy = – 5.0 m/s and t = 2.0 s  h = (– 5.0) × 2.0 + 1/2 × 10 × (2.0)2 = – 10 + 20 = 10 meter.