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Page 1 14. Vertical velocity of balloon (+ bag) = ´ 12 5 18 m/s = 10 3 m/s Horizontal velocity of balloon (+ bag) = Wind velocity = = ´ 20 20 5 18 km h / m/s = 50 9 m/s = 5.55 m/s \ tan a = 12 20 . i.e., sin a = 0.51 Bag is released at point A. Let t be time, the bag takes from A to reach ground. Using, s ut at = + 1 2 2 ( ) sin ( )  = æ è ç ö ø ÷ +  50 10 3 1 2 2 a t g t i.e., 5 50 0 2 t t   = 1.7 \ t = +   ´ ´  ´ 1.7 1.7 4 ( ) ( ) 2 5 50 2 5 = 3.37 s Vertical velocity of bag when it strikes ground v B =  + 10 3 10 ( )( ) 3.37 =37.03 m/s v w =5.55 m/s \ Velocity of bag with which it strikes ground v v v B net = + 2 2 w = 37.44 m/s 15. T u g =  2 sin ( ) cos a b b = ´ ´ °  ° ° 2 20 2 45 30 10 30 sin( ) cos =1.69 s R u g =  2 2 2 sin( )cos cos a b a b = ´ ´ ° ´ ° ° 2 20 2 15 45 10 30 2 2 ( ) sin cos cos =39 m 16. T u g = + 2 sin( ) cos a b b = ´ ´ ° + ° ° 2 20 2 45 30 10 30 sin( ) cos =6.31 s R u g = + + 2 2 2 cos [sin ( ) sin ] b a b b = ° ° + ° + ° ( ) cos [sin ( ) sin ] 20 2 10 30 90 30 30 2 2 = 145.71 m 17. T u g = + 2 sin( ) cos a b b = 2u g sin cos b b (Q a = ° 0 ) = 2u g tanb = ´ ° 2 20 10 30 tan =2.31 s R u g = + + 2 2 2 cos [sin ( ) sin ] b a b b = u g 2 2 2 ( sin ) cos b b [as a = ° 0 ] = uT cosb = ´ ° 20 30 2.31 cos = 53.33 m 18. R u g = + + 2 2 2 cos [sin ( ) sin ] b a b b = +  + u g 2 2 2 cos [sin{ ( ) } sin ] b a b b b =  + u g 2 2 cos [sin( ) sin ] q p q q [Q ( ) a b p + = 2 ] or R u g = 2 2 tanq q sec 60  Mechanics1 a a O 50 m A 20 km/h v = w 12 km/h Page 2 14. Vertical velocity of balloon (+ bag) = ´ 12 5 18 m/s = 10 3 m/s Horizontal velocity of balloon (+ bag) = Wind velocity = = ´ 20 20 5 18 km h / m/s = 50 9 m/s = 5.55 m/s \ tan a = 12 20 . i.e., sin a = 0.51 Bag is released at point A. Let t be time, the bag takes from A to reach ground. Using, s ut at = + 1 2 2 ( ) sin ( )  = æ è ç ö ø ÷ +  50 10 3 1 2 2 a t g t i.e., 5 50 0 2 t t   = 1.7 \ t = +   ´ ´  ´ 1.7 1.7 4 ( ) ( ) 2 5 50 2 5 = 3.37 s Vertical velocity of bag when it strikes ground v B =  + 10 3 10 ( )( ) 3.37 =37.03 m/s v w =5.55 m/s \ Velocity of bag with which it strikes ground v v v B net = + 2 2 w = 37.44 m/s 15. T u g =  2 sin ( ) cos a b b = ´ ´ °  ° ° 2 20 2 45 30 10 30 sin( ) cos =1.69 s R u g =  2 2 2 sin( )cos cos a b a b = ´ ´ ° ´ ° ° 2 20 2 15 45 10 30 2 2 ( ) sin cos cos =39 m 16. T u g = + 2 sin( ) cos a b b = ´ ´ ° + ° ° 2 20 2 45 30 10 30 sin( ) cos =6.31 s R u g = + + 2 2 2 cos [sin ( ) sin ] b a b b = ° ° + ° + ° ( ) cos [sin ( ) sin ] 20 2 10 30 90 30 30 2 2 = 145.71 m 17. T u g = + 2 sin( ) cos a b b = 2u g sin cos b b (Q a = ° 0 ) = 2u g tanb = ´ ° 2 20 10 30 tan =2.31 s R u g = + + 2 2 2 cos [sin ( ) sin ] b a b b = u g 2 2 2 ( sin ) cos b b [as a = ° 0 ] = uT cosb = ´ ° 20 30 2.31 cos = 53.33 m 18. R u g = + + 2 2 2 cos [sin ( ) sin ] b a b b = +  + u g 2 2 2 cos [sin{ ( ) } sin ] b a b b b =  + u g 2 2 cos [sin( ) sin ] q p q q [Q ( ) a b p + = 2 ] or R u g = 2 2 tanq q sec 60  Mechanics1 a a O 50 m A 20 km/h v = w 12 km/h 19. (a) Acceleration of particle 1 w.r.t. that of particle 2 =    ( ) ( ) g g =0 (b) Initial velocity of 1st particle = 20 j ^ m/s Initial velocity of 2nd particle = ° + ° ( cos sin ) ^ ^ 20 2 45 20 2 45 i j m/s = + ( ) ^ ^ 20 20 i j m/s \ Initial velocity of 1st particle w.r.t. that of 2nd particle =  + [( ) ( )] ^ ^ ^ 20 20 20 j i j m/s = 20i ^ m/s =20 m/s (downward) (c) Horizontal velocity of 1st particle =0 m/s Horizontal velocity of 2nd particle = 20 i ^ m/s \ Horizontal velocity of 1st particle w.r.t. that of 2nd particle =  0 20 ( ) ^ i =  20 i ^ m/s Relative displacement of 1st particle w.r.t. 2nd particle at t = 2 s =  ´ 20 2 i ^ =  40 i ^ m/s \ Distance between the particles at t = 2 s = 40 m 20. (a) As observed by passenger Vertical acceleration of stone =  = g g 0 Horizontal velocity of stone =  = v v 0 \ Path of the stone will be a straight line (downwards). (b)As observed by man standing on ground Vertical acceleration of stone = g Horizontal velocity of stone = v \ Path of the stone will be parabolic. 21. (a) g g a eff =   ( ) = + g a = + 10 1 = 11 m/s 2 T u g = 2 sin q eff = ´ ´ ° 2 2 30 11 sin = 0.18 s (b)Dotted path [(in lift) acceleration upwards] Full line path [In lift at rest or moving with constant velocity upwards or downwards]. (c)If lift is moving downward with acceleration g. g g g eff =  =0 22. Horizontal motion : x u 1 1 1 = cos q and x u 2 2 = cos q \ u u 1 1 2 2 20 cos cos q q + = …(i) Projectile Motion 61 q u 2 a m/s w m/s q u q Path of particle u Page 3 14. Vertical velocity of balloon (+ bag) = ´ 12 5 18 m/s = 10 3 m/s Horizontal velocity of balloon (+ bag) = Wind velocity = = ´ 20 20 5 18 km h / m/s = 50 9 m/s = 5.55 m/s \ tan a = 12 20 . i.e., sin a = 0.51 Bag is released at point A. Let t be time, the bag takes from A to reach ground. Using, s ut at = + 1 2 2 ( ) sin ( )  = æ è ç ö ø ÷ +  50 10 3 1 2 2 a t g t i.e., 5 50 0 2 t t   = 1.7 \ t = +   ´ ´  ´ 1.7 1.7 4 ( ) ( ) 2 5 50 2 5 = 3.37 s Vertical velocity of bag when it strikes ground v B =  + 10 3 10 ( )( ) 3.37 =37.03 m/s v w =5.55 m/s \ Velocity of bag with which it strikes ground v v v B net = + 2 2 w = 37.44 m/s 15. T u g =  2 sin ( ) cos a b b = ´ ´ °  ° ° 2 20 2 45 30 10 30 sin( ) cos =1.69 s R u g =  2 2 2 sin( )cos cos a b a b = ´ ´ ° ´ ° ° 2 20 2 15 45 10 30 2 2 ( ) sin cos cos =39 m 16. T u g = + 2 sin( ) cos a b b = ´ ´ ° + ° ° 2 20 2 45 30 10 30 sin( ) cos =6.31 s R u g = + + 2 2 2 cos [sin ( ) sin ] b a b b = ° ° + ° + ° ( ) cos [sin ( ) sin ] 20 2 10 30 90 30 30 2 2 = 145.71 m 17. T u g = + 2 sin( ) cos a b b = 2u g sin cos b b (Q a = ° 0 ) = 2u g tanb = ´ ° 2 20 10 30 tan =2.31 s R u g = + + 2 2 2 cos [sin ( ) sin ] b a b b = u g 2 2 2 ( sin ) cos b b [as a = ° 0 ] = uT cosb = ´ ° 20 30 2.31 cos = 53.33 m 18. R u g = + + 2 2 2 cos [sin ( ) sin ] b a b b = +  + u g 2 2 2 cos [sin{ ( ) } sin ] b a b b b =  + u g 2 2 cos [sin( ) sin ] q p q q [Q ( ) a b p + = 2 ] or R u g = 2 2 tanq q sec 60  Mechanics1 a a O 50 m A 20 km/h v = w 12 km/h 19. (a) Acceleration of particle 1 w.r.t. that of particle 2 =    ( ) ( ) g g =0 (b) Initial velocity of 1st particle = 20 j ^ m/s Initial velocity of 2nd particle = ° + ° ( cos sin ) ^ ^ 20 2 45 20 2 45 i j m/s = + ( ) ^ ^ 20 20 i j m/s \ Initial velocity of 1st particle w.r.t. that of 2nd particle =  + [( ) ( )] ^ ^ ^ 20 20 20 j i j m/s = 20i ^ m/s =20 m/s (downward) (c) Horizontal velocity of 1st particle =0 m/s Horizontal velocity of 2nd particle = 20 i ^ m/s \ Horizontal velocity of 1st particle w.r.t. that of 2nd particle =  0 20 ( ) ^ i =  20 i ^ m/s Relative displacement of 1st particle w.r.t. 2nd particle at t = 2 s =  ´ 20 2 i ^ =  40 i ^ m/s \ Distance between the particles at t = 2 s = 40 m 20. (a) As observed by passenger Vertical acceleration of stone =  = g g 0 Horizontal velocity of stone =  = v v 0 \ Path of the stone will be a straight line (downwards). (b)As observed by man standing on ground Vertical acceleration of stone = g Horizontal velocity of stone = v \ Path of the stone will be parabolic. 21. (a) g g a eff =   ( ) = + g a = + 10 1 = 11 m/s 2 T u g = 2 sin q eff = ´ ´ ° 2 2 30 11 sin = 0.18 s (b)Dotted path [(in lift) acceleration upwards] Full line path [In lift at rest or moving with constant velocity upwards or downwards]. (c)If lift is moving downward with acceleration g. g g g eff =  =0 22. Horizontal motion : x u 1 1 1 = cos q and x u 2 2 = cos q \ u u 1 1 2 2 20 cos cos q q + = …(i) Projectile Motion 61 q u 2 a m/s w m/s q u q Path of particle u Vertical motion : 20 1 2 30 1 1 2 + +  = ( sin ) ( ) u t g t q + +  ( sin ) ( ) u t g t 2 2 2 1 2 q or ( sin sin ) u u t 1 1 2 2 10 q q  = …(ii) Objective Questions (Level 1) 1. v i j ® = + 3 4 ^ ^ and F i j ® =  4 3 ^ ^ v F i j i j ® ® × = + ×  ( ) ( ) ^ ^ ^ ^ 3 4 4 3 =  = 12 12 0 \ F v ® ® ^ Path of the particle is circular. Option (c) is correct. 2. Projectile motion is uniformly accelerated everywhere even at the highest point. Option (a) correct. Option (b) incorrect. At the highest point acceleration is perpendicular to velocity. Option (c) incorrect. 3. For range to be maximum q = ° 45 i.e., u u x = ° = cos 45 1 2 or v u x = 2 ( Q v u x x = ) = 20 2 = 14.14 m/s = 14 m/s (approx) Option (b) is correct. 4. H (maximum height) = u g 2 2 2 sin a \ H u g 1 2 2 2 = sin q and H u g 2 2 2 90 = °  sin ( ) q = u g 2 2 cos q Thus, H H 1 2 2 2 = sin cos q q Option (c) is correct. 5. Equation to trajectory is Y x gx u =  tan cos sin q q q 1 2 2 2 Y x x R =  tanq 2 Area = ò Y dx R 0 =  æ è ç ö ø ÷ ò x x R dx R tanq 2 0 A x x R R =  é ë ê ù û ú 2 3 0 2 3 tanq =  R R 2 3 2 3 tanq =  é ë ê ù û ú R 2 2 1 3 tanq 62  Mechanics1 Y v F X i j ® ® u 2 q 1 x 2 x 1 d = 20 m 20 1 u 1 q 2 t Page 4 14. Vertical velocity of balloon (+ bag) = ´ 12 5 18 m/s = 10 3 m/s Horizontal velocity of balloon (+ bag) = Wind velocity = = ´ 20 20 5 18 km h / m/s = 50 9 m/s = 5.55 m/s \ tan a = 12 20 . i.e., sin a = 0.51 Bag is released at point A. Let t be time, the bag takes from A to reach ground. Using, s ut at = + 1 2 2 ( ) sin ( )  = æ è ç ö ø ÷ +  50 10 3 1 2 2 a t g t i.e., 5 50 0 2 t t   = 1.7 \ t = +   ´ ´  ´ 1.7 1.7 4 ( ) ( ) 2 5 50 2 5 = 3.37 s Vertical velocity of bag when it strikes ground v B =  + 10 3 10 ( )( ) 3.37 =37.03 m/s v w =5.55 m/s \ Velocity of bag with which it strikes ground v v v B net = + 2 2 w = 37.44 m/s 15. T u g =  2 sin ( ) cos a b b = ´ ´ °  ° ° 2 20 2 45 30 10 30 sin( ) cos =1.69 s R u g =  2 2 2 sin( )cos cos a b a b = ´ ´ ° ´ ° ° 2 20 2 15 45 10 30 2 2 ( ) sin cos cos =39 m 16. T u g = + 2 sin( ) cos a b b = ´ ´ ° + ° ° 2 20 2 45 30 10 30 sin( ) cos =6.31 s R u g = + + 2 2 2 cos [sin ( ) sin ] b a b b = ° ° + ° + ° ( ) cos [sin ( ) sin ] 20 2 10 30 90 30 30 2 2 = 145.71 m 17. T u g = + 2 sin( ) cos a b b = 2u g sin cos b b (Q a = ° 0 ) = 2u g tanb = ´ ° 2 20 10 30 tan =2.31 s R u g = + + 2 2 2 cos [sin ( ) sin ] b a b b = u g 2 2 2 ( sin ) cos b b [as a = ° 0 ] = uT cosb = ´ ° 20 30 2.31 cos = 53.33 m 18. R u g = + + 2 2 2 cos [sin ( ) sin ] b a b b = +  + u g 2 2 2 cos [sin{ ( ) } sin ] b a b b b =  + u g 2 2 cos [sin( ) sin ] q p q q [Q ( ) a b p + = 2 ] or R u g = 2 2 tanq q sec 60  Mechanics1 a a O 50 m A 20 km/h v = w 12 km/h 19. (a) Acceleration of particle 1 w.r.t. that of particle 2 =    ( ) ( ) g g =0 (b) Initial velocity of 1st particle = 20 j ^ m/s Initial velocity of 2nd particle = ° + ° ( cos sin ) ^ ^ 20 2 45 20 2 45 i j m/s = + ( ) ^ ^ 20 20 i j m/s \ Initial velocity of 1st particle w.r.t. that of 2nd particle =  + [( ) ( )] ^ ^ ^ 20 20 20 j i j m/s = 20i ^ m/s =20 m/s (downward) (c) Horizontal velocity of 1st particle =0 m/s Horizontal velocity of 2nd particle = 20 i ^ m/s \ Horizontal velocity of 1st particle w.r.t. that of 2nd particle =  0 20 ( ) ^ i =  20 i ^ m/s Relative displacement of 1st particle w.r.t. 2nd particle at t = 2 s =  ´ 20 2 i ^ =  40 i ^ m/s \ Distance between the particles at t = 2 s = 40 m 20. (a) As observed by passenger Vertical acceleration of stone =  = g g 0 Horizontal velocity of stone =  = v v 0 \ Path of the stone will be a straight line (downwards). (b)As observed by man standing on ground Vertical acceleration of stone = g Horizontal velocity of stone = v \ Path of the stone will be parabolic. 21. (a) g g a eff =   ( ) = + g a = + 10 1 = 11 m/s 2 T u g = 2 sin q eff = ´ ´ ° 2 2 30 11 sin = 0.18 s (b)Dotted path [(in lift) acceleration upwards] Full line path [In lift at rest or moving with constant velocity upwards or downwards]. (c)If lift is moving downward with acceleration g. g g g eff =  =0 22. Horizontal motion : x u 1 1 1 = cos q and x u 2 2 = cos q \ u u 1 1 2 2 20 cos cos q q + = …(i) Projectile Motion 61 q u 2 a m/s w m/s q u q Path of particle u Vertical motion : 20 1 2 30 1 1 2 + +  = ( sin ) ( ) u t g t q + +  ( sin ) ( ) u t g t 2 2 2 1 2 q or ( sin sin ) u u t 1 1 2 2 10 q q  = …(ii) Objective Questions (Level 1) 1. v i j ® = + 3 4 ^ ^ and F i j ® =  4 3 ^ ^ v F i j i j ® ® × = + ×  ( ) ( ) ^ ^ ^ ^ 3 4 4 3 =  = 12 12 0 \ F v ® ® ^ Path of the particle is circular. Option (c) is correct. 2. Projectile motion is uniformly accelerated everywhere even at the highest point. Option (a) correct. Option (b) incorrect. At the highest point acceleration is perpendicular to velocity. Option (c) incorrect. 3. For range to be maximum q = ° 45 i.e., u u x = ° = cos 45 1 2 or v u x = 2 ( Q v u x x = ) = 20 2 = 14.14 m/s = 14 m/s (approx) Option (b) is correct. 4. H (maximum height) = u g 2 2 2 sin a \ H u g 1 2 2 2 = sin q and H u g 2 2 2 90 = °  sin ( ) q = u g 2 2 cos q Thus, H H 1 2 2 2 = sin cos q q Option (c) is correct. 5. Equation to trajectory is Y x gx u =  tan cos sin q q q 1 2 2 2 Y x x R =  tanq 2 Area = ò Y dx R 0 =  æ è ç ö ø ÷ ò x x R dx R tanq 2 0 A x x R R =  é ë ê ù û ú 2 3 0 2 3 tanq =  R R 2 3 2 3 tanq =  é ë ê ù û ú R 2 2 1 3 tanq 62  Mechanics1 Y v F X i j ® ® u 2 q 1 x 2 x 1 d = 20 m 20 1 u 1 q 2 t =  é ë ê ù û ú 4 2 1 3 0 4 2 2 2 v g sin cos tan q q q =  é ë ê ù û ú 4 2 3 0 4 2 3 2 2 v g sin cos sin cos q q q q =  2 3 3 2 0 4 2 2 2 2 v g [ sin cos sin cos ] q q q q 6. v u cos cos f= q or v u = ° ° cos cos 60 30 or v u = 3 KE at B mv = 1 2 2 = × 1 2 3 2 m u = K 3 Q 1 2 2 mu K = æ è ç ö ø ÷ \ Option (b) is correct. 7. u g u g 2 2 2 1 2 sin q = æ è ç ö ø ÷ or sin 2 1 2 q = or 2 30 q = ° \ q = ° 15 Option (a) is correct. 8. T u g 1 2 = sin q \ T u g 2 2 90 = °  sin ( ) q = 2u g cosq Thus, TT g u g 1 2 2 2 2 = sin cos q q or TT R g 1 2 2 = Þ R gTT = 1 2 1 2 Option (d) is correct. 9. R max = 1.6 m u g 2 = 1.6 Þ u = 4 m/s \ T u g = ° 2 45 sin = 4 2 10 Number of jumps = = 10 2 10 2 4 2 10 T / = 25 \ Grass hopper would go = ´ 25 1.6 m i.e., 40 m. Option (d) is correct. 10. Av. velocity= Displacement time = æ è ç ö ø ÷ + 1 2 2 2 2 T R H / = æ è ç ö ø ÷ + æ è ç ö ø ÷ 1 2 2 2 2 2 2 u g u g u g sin sin cos sin q q q q = + u cos sin 2 2 4 q q = + u 2 1 3 2 cos q Option (b) is correct. 11. d R u T 2 2 2 2 = + = æ è ç ö ø ÷ + ° æ è ç ö ø ÷ u g u u g 2 2 2 2 2 45 sin = + u g u g 4 2 4 2 2 = 3 4 2 u g Projectile Motion 63 q u A f v b 90° 45° u 2 R = u /g T Ball uT a T Page 5 14. Vertical velocity of balloon (+ bag) = ´ 12 5 18 m/s = 10 3 m/s Horizontal velocity of balloon (+ bag) = Wind velocity = = ´ 20 20 5 18 km h / m/s = 50 9 m/s = 5.55 m/s \ tan a = 12 20 . i.e., sin a = 0.51 Bag is released at point A. Let t be time, the bag takes from A to reach ground. Using, s ut at = + 1 2 2 ( ) sin ( )  = æ è ç ö ø ÷ +  50 10 3 1 2 2 a t g t i.e., 5 50 0 2 t t   = 1.7 \ t = +   ´ ´  ´ 1.7 1.7 4 ( ) ( ) 2 5 50 2 5 = 3.37 s Vertical velocity of bag when it strikes ground v B =  + 10 3 10 ( )( ) 3.37 =37.03 m/s v w =5.55 m/s \ Velocity of bag with which it strikes ground v v v B net = + 2 2 w = 37.44 m/s 15. T u g =  2 sin ( ) cos a b b = ´ ´ °  ° ° 2 20 2 45 30 10 30 sin( ) cos =1.69 s R u g =  2 2 2 sin( )cos cos a b a b = ´ ´ ° ´ ° ° 2 20 2 15 45 10 30 2 2 ( ) sin cos cos =39 m 16. T u g = + 2 sin( ) cos a b b = ´ ´ ° + ° ° 2 20 2 45 30 10 30 sin( ) cos =6.31 s R u g = + + 2 2 2 cos [sin ( ) sin ] b a b b = ° ° + ° + ° ( ) cos [sin ( ) sin ] 20 2 10 30 90 30 30 2 2 = 145.71 m 17. T u g = + 2 sin( ) cos a b b = 2u g sin cos b b (Q a = ° 0 ) = 2u g tanb = ´ ° 2 20 10 30 tan =2.31 s R u g = + + 2 2 2 cos [sin ( ) sin ] b a b b = u g 2 2 2 ( sin ) cos b b [as a = ° 0 ] = uT cosb = ´ ° 20 30 2.31 cos = 53.33 m 18. R u g = + + 2 2 2 cos [sin ( ) sin ] b a b b = +  + u g 2 2 2 cos [sin{ ( ) } sin ] b a b b b =  + u g 2 2 cos [sin( ) sin ] q p q q [Q ( ) a b p + = 2 ] or R u g = 2 2 tanq q sec 60  Mechanics1 a a O 50 m A 20 km/h v = w 12 km/h 19. (a) Acceleration of particle 1 w.r.t. that of particle 2 =    ( ) ( ) g g =0 (b) Initial velocity of 1st particle = 20 j ^ m/s Initial velocity of 2nd particle = ° + ° ( cos sin ) ^ ^ 20 2 45 20 2 45 i j m/s = + ( ) ^ ^ 20 20 i j m/s \ Initial velocity of 1st particle w.r.t. that of 2nd particle =  + [( ) ( )] ^ ^ ^ 20 20 20 j i j m/s = 20i ^ m/s =20 m/s (downward) (c) Horizontal velocity of 1st particle =0 m/s Horizontal velocity of 2nd particle = 20 i ^ m/s \ Horizontal velocity of 1st particle w.r.t. that of 2nd particle =  0 20 ( ) ^ i =  20 i ^ m/s Relative displacement of 1st particle w.r.t. 2nd particle at t = 2 s =  ´ 20 2 i ^ =  40 i ^ m/s \ Distance between the particles at t = 2 s = 40 m 20. (a) As observed by passenger Vertical acceleration of stone =  = g g 0 Horizontal velocity of stone =  = v v 0 \ Path of the stone will be a straight line (downwards). (b)As observed by man standing on ground Vertical acceleration of stone = g Horizontal velocity of stone = v \ Path of the stone will be parabolic. 21. (a) g g a eff =   ( ) = + g a = + 10 1 = 11 m/s 2 T u g = 2 sin q eff = ´ ´ ° 2 2 30 11 sin = 0.18 s (b)Dotted path [(in lift) acceleration upwards] Full line path [In lift at rest or moving with constant velocity upwards or downwards]. (c)If lift is moving downward with acceleration g. g g g eff =  =0 22. Horizontal motion : x u 1 1 1 = cos q and x u 2 2 = cos q \ u u 1 1 2 2 20 cos cos q q + = …(i) Projectile Motion 61 q u 2 a m/s w m/s q u q Path of particle u Vertical motion : 20 1 2 30 1 1 2 + +  = ( sin ) ( ) u t g t q + +  ( sin ) ( ) u t g t 2 2 2 1 2 q or ( sin sin ) u u t 1 1 2 2 10 q q  = …(ii) Objective Questions (Level 1) 1. v i j ® = + 3 4 ^ ^ and F i j ® =  4 3 ^ ^ v F i j i j ® ® × = + ×  ( ) ( ) ^ ^ ^ ^ 3 4 4 3 =  = 12 12 0 \ F v ® ® ^ Path of the particle is circular. Option (c) is correct. 2. Projectile motion is uniformly accelerated everywhere even at the highest point. Option (a) correct. Option (b) incorrect. At the highest point acceleration is perpendicular to velocity. Option (c) incorrect. 3. For range to be maximum q = ° 45 i.e., u u x = ° = cos 45 1 2 or v u x = 2 ( Q v u x x = ) = 20 2 = 14.14 m/s = 14 m/s (approx) Option (b) is correct. 4. H (maximum height) = u g 2 2 2 sin a \ H u g 1 2 2 2 = sin q and H u g 2 2 2 90 = °  sin ( ) q = u g 2 2 cos q Thus, H H 1 2 2 2 = sin cos q q Option (c) is correct. 5. Equation to trajectory is Y x gx u =  tan cos sin q q q 1 2 2 2 Y x x R =  tanq 2 Area = ò Y dx R 0 =  æ è ç ö ø ÷ ò x x R dx R tanq 2 0 A x x R R =  é ë ê ù û ú 2 3 0 2 3 tanq =  R R 2 3 2 3 tanq =  é ë ê ù û ú R 2 2 1 3 tanq 62  Mechanics1 Y v F X i j ® ® u 2 q 1 x 2 x 1 d = 20 m 20 1 u 1 q 2 t =  é ë ê ù û ú 4 2 1 3 0 4 2 2 2 v g sin cos tan q q q =  é ë ê ù û ú 4 2 3 0 4 2 3 2 2 v g sin cos sin cos q q q q =  2 3 3 2 0 4 2 2 2 2 v g [ sin cos sin cos ] q q q q 6. v u cos cos f= q or v u = ° ° cos cos 60 30 or v u = 3 KE at B mv = 1 2 2 = × 1 2 3 2 m u = K 3 Q 1 2 2 mu K = æ è ç ö ø ÷ \ Option (b) is correct. 7. u g u g 2 2 2 1 2 sin q = æ è ç ö ø ÷ or sin 2 1 2 q = or 2 30 q = ° \ q = ° 15 Option (a) is correct. 8. T u g 1 2 = sin q \ T u g 2 2 90 = °  sin ( ) q = 2u g cosq Thus, TT g u g 1 2 2 2 2 = sin cos q q or TT R g 1 2 2 = Þ R gTT = 1 2 1 2 Option (d) is correct. 9. R max = 1.6 m u g 2 = 1.6 Þ u = 4 m/s \ T u g = ° 2 45 sin = 4 2 10 Number of jumps = = 10 2 10 2 4 2 10 T / = 25 \ Grass hopper would go = ´ 25 1.6 m i.e., 40 m. Option (d) is correct. 10. Av. velocity= Displacement time = æ è ç ö ø ÷ + 1 2 2 2 2 T R H / = æ è ç ö ø ÷ + æ è ç ö ø ÷ 1 2 2 2 2 2 2 u g u g u g sin sin cos sin q q q q = + u cos sin 2 2 4 q q = + u 2 1 3 2 cos q Option (b) is correct. 11. d R u T 2 2 2 2 = + = æ è ç ö ø ÷ + ° æ è ç ö ø ÷ u g u u g 2 2 2 2 2 45 sin = + u g u g 4 2 4 2 2 = 3 4 2 u g Projectile Motion 63 q u A f v b 90° 45° u 2 R = u /g T Ball uT a T \ d u g = 2 3 = 30 10 3 2 = 90 3 m Option (b) is correct. 12. At maximum height dy dt = 0 i.e., d dt t t ( ) 10 0 2  = or 10 2 0  = t or t =5 s \ Maximum height attained =  10 5 5 2 ( ) = 25 m Option (d) is correct. 13. R u g = + + 2 2 2 cos [sin ( ) sin ] b a b b As a = 0 (according to question) R u g = 2 2 2 cos [ sin ] b b = ´ ° ´ ° ( ) sin cos 50 2 30 10 30 2 2 = 1000 3 m Option (b) is correct. 14. First particle : H max =102 m \ u g 2 2 2 102 sin a = Þ u g 2 2 102 2 60 = ´ ° sin [As a p = = ° 3 60 ] Second particle : Range of the second particle will be equal to that of particle if, u g u g 2 2 2 2 sin sin f = a sin sin 2 3 f = a 2 2 f =  p a or f =  p a 2 =  p p 2 3 = = ° p 6 30 \ Maximum height attained by second particle = f u g 2 2 2 sin = ´ ° ´ ° 102 2 60 30 2 2 2 g g sin sin = ´ 102 1 4 3 4 / / =34 m Option (d) is correct. 15. s ut at = + 1 2 2 \ ( ) ( sin ) ( )  = ° +  70 50 30 1 2 10 2 t t or 5 25 70 0 2 t t   = or t t 2 5 14 0   = ( ) ( ) t t  + = 7 2 0 t = 7 s ( 2 s not possible) Option (c) is correct. 16. Initial separation x R T = +5 = + ( cos ) u T T a 5 = + T u [( cos ) ] a 5 = + 2 5 u g u sin [( cos ) ] a a = ´ ´ ´ æ è ç ö ø ÷ + é ë ê ù û ú 2 80 4 5 10 80 3 5 5 =256 m Option (d) is correct. 64  Mechanics1 53° T 5T R x b a u a = –g RRead More
1. What is projectile motion? 
2. How is the trajectory of a projectile determined? 
3. How can we calculate the maximum height reached by a projectile? 
4. What is the range of a projectile? 
5. How does air resistance affect projectile motion? 

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