Page 1 Introductory Exercise 4.1 1. A particle projected at any angle with horizontal will always move in a plane and thus projectile motion is a 2-dimensional motion. The statement is thus false. 2. At high speed the projectile may go to a place where acceleration due to gravity has some different value and as such the motion may not be uniform accelerated. The statement is thus true. 3. See article 4.1. 4. u = 40 2 m/s, q = ° 45 As horizontal acceleration would be zero. v u u x x = = = cos q 40 m/s s u t u t x x = = = ( cos ) q 80 m A : position of particle at time = 0. B : position of particle at time = t. As vertical acceleration would be - g v u gt y y = - = = u gt sinq = - 40 20 =20 m/s s u t gt y y = × - 1 2 2 = - ´ ´ = 80 1 2 10 2 60 2 m \ v v v x y = + 2 2 = + 40 20 2 2 = 20 5 m/s tan f = = v v y x 20 40 i.e., f = æ è ç ö ø ÷ - tan 1 1 2 s s s x y = + 2 2 = + 80 60 2 2 = 100 m tan a = s s y x = 60 80 i.e., a = æ è ç ö ø ÷ - tan 1 3 4 5. s u t g t y y = + - 1 2 2 ( ) or s u t gt y = - ( sin ) q 1 2 2 or 15 20 1 2 2 = - t gt or t t 2 4 3 0 = + = i.e., t = 1 s and 3 s 6. See figure to the answer to question no. 4. u = 40 m/s, q = ° 60 \ u x = ° = 40 60 20 cos m/s Projectile Motion 4 s y g f v v y v x u x s x A a u u y B Page 2 Introductory Exercise 4.1 1. A particle projected at any angle with horizontal will always move in a plane and thus projectile motion is a 2-dimensional motion. The statement is thus false. 2. At high speed the projectile may go to a place where acceleration due to gravity has some different value and as such the motion may not be uniform accelerated. The statement is thus true. 3. See article 4.1. 4. u = 40 2 m/s, q = ° 45 As horizontal acceleration would be zero. v u u x x = = = cos q 40 m/s s u t u t x x = = = ( cos ) q 80 m A : position of particle at time = 0. B : position of particle at time = t. As vertical acceleration would be - g v u gt y y = - = = u gt sinq = - 40 20 =20 m/s s u t gt y y = × - 1 2 2 = - ´ ´ = 80 1 2 10 2 60 2 m \ v v v x y = + 2 2 = + 40 20 2 2 = 20 5 m/s tan f = = v v y x 20 40 i.e., f = æ è ç ö ø ÷ - tan 1 1 2 s s s x y = + 2 2 = + 80 60 2 2 = 100 m tan a = s s y x = 60 80 i.e., a = æ è ç ö ø ÷ - tan 1 3 4 5. s u t g t y y = + - 1 2 2 ( ) or s u t gt y = - ( sin ) q 1 2 2 or 15 20 1 2 2 = - t gt or t t 2 4 3 0 = + = i.e., t = 1 s and 3 s 6. See figure to the answer to question no. 4. u = 40 m/s, q = ° 60 \ u x = ° = 40 60 20 cos m/s Projectile Motion 4 s y g f v v y v x u x s x A a u u y B Thus, v u x x = = 20 m/s As f = ° 45 tan f = = v v y x 1 \ y v y x = = 20 m/s Thus, v v v v x y x = + = 2 2 2 = 20 2 m/s Before reaching highest point v u g t y y = + - ( ) \ 20 40 60 10 = ° - sin t or 2 2 3 = - t Þ t = - 2 3 1 ( ) s After attaining highest point - = - 20 40 3 10t i.e., - = - 2 2 3 t or t = + 2 3 1 ( ) s 7. Average velocity = Range Time of flight = u g u g 2 2 sin 2a a / / sin = ´ 2 2 2 u g g u sin cos sin a a a = ucos a 8. Change in velocity = - - + ( sin ) ( sin ) u u a a = -2usina =2usina (downward) 9. Formulae for R T , and H max will be same if the projection point and the point where the particle lands are same and lie on a horizontal line. 10. r i j ® = + - [ ( ) ] ^ ^ 3 4 5 2 t t t m y-coordinate will be zero when 4 5 0 2 t t - = i.e., t = 0 s, 4 5 s t = 0 belongs to the initial point of projection of the particle. \ r i ® = 0 ^ m i.e., x = 0 m At t = 0.8 s, r i ® = 2.4 ^ m i.e., x = 2.4 m 11. v u x x = = 10 m/s =10 m/s v u x x = = 10 m/s v v x = f cos Þ cosf= v v x = 10 v = 10 10 [as, v = 20 2 (given)] =1 i.e., f = ° 0 \ Speed will be half of its initial value at the highest point where f = ° 0 . Thus, t u g = sin q = ° = 20 60 10 3 sin s 52 | Mechanics-1 a a u â€“ u sin a u sin a t = T u cos a + u sin a u 60° u = 20 m/s v x v f u = 20 cos 60° x Page 3 Introductory Exercise 4.1 1. A particle projected at any angle with horizontal will always move in a plane and thus projectile motion is a 2-dimensional motion. The statement is thus false. 2. At high speed the projectile may go to a place where acceleration due to gravity has some different value and as such the motion may not be uniform accelerated. The statement is thus true. 3. See article 4.1. 4. u = 40 2 m/s, q = ° 45 As horizontal acceleration would be zero. v u u x x = = = cos q 40 m/s s u t u t x x = = = ( cos ) q 80 m A : position of particle at time = 0. B : position of particle at time = t. As vertical acceleration would be - g v u gt y y = - = = u gt sinq = - 40 20 =20 m/s s u t gt y y = × - 1 2 2 = - ´ ´ = 80 1 2 10 2 60 2 m \ v v v x y = + 2 2 = + 40 20 2 2 = 20 5 m/s tan f = = v v y x 20 40 i.e., f = æ è ç ö ø ÷ - tan 1 1 2 s s s x y = + 2 2 = + 80 60 2 2 = 100 m tan a = s s y x = 60 80 i.e., a = æ è ç ö ø ÷ - tan 1 3 4 5. s u t g t y y = + - 1 2 2 ( ) or s u t gt y = - ( sin ) q 1 2 2 or 15 20 1 2 2 = - t gt or t t 2 4 3 0 = + = i.e., t = 1 s and 3 s 6. See figure to the answer to question no. 4. u = 40 m/s, q = ° 60 \ u x = ° = 40 60 20 cos m/s Projectile Motion 4 s y g f v v y v x u x s x A a u u y B Thus, v u x x = = 20 m/s As f = ° 45 tan f = = v v y x 1 \ y v y x = = 20 m/s Thus, v v v v x y x = + = 2 2 2 = 20 2 m/s Before reaching highest point v u g t y y = + - ( ) \ 20 40 60 10 = ° - sin t or 2 2 3 = - t Þ t = - 2 3 1 ( ) s After attaining highest point - = - 20 40 3 10t i.e., - = - 2 2 3 t or t = + 2 3 1 ( ) s 7. Average velocity = Range Time of flight = u g u g 2 2 sin 2a a / / sin = ´ 2 2 2 u g g u sin cos sin a a a = ucos a 8. Change in velocity = - - + ( sin ) ( sin ) u u a a = -2usina =2usina (downward) 9. Formulae for R T , and H max will be same if the projection point and the point where the particle lands are same and lie on a horizontal line. 10. r i j ® = + - [ ( ) ] ^ ^ 3 4 5 2 t t t m y-coordinate will be zero when 4 5 0 2 t t - = i.e., t = 0 s, 4 5 s t = 0 belongs to the initial point of projection of the particle. \ r i ® = 0 ^ m i.e., x = 0 m At t = 0.8 s, r i ® = 2.4 ^ m i.e., x = 2.4 m 11. v u x x = = 10 m/s =10 m/s v u x x = = 10 m/s v v x = f cos Þ cosf= v v x = 10 v = 10 10 [as, v = 20 2 (given)] =1 i.e., f = ° 0 \ Speed will be half of its initial value at the highest point where f = ° 0 . Thus, t u g = sin q = ° = 20 60 10 3 sin s 52 | Mechanics-1 a a u â€“ u sin a u sin a t = T u cos a + u sin a u 60° u = 20 m/s v x v f u = 20 cos 60° x Introductory Exercise 4.2 1. Time of flight T u g = - 2 sin ( ) cos a b b = ´ ° - ° ° ( sin( ) cos 2 10 60 30 10 30 = 2 3 s Using, v u at = + v u u x x = = ° cos 60 = ´ = 10 1 2 5 m/s v u g T y y = + - ( ) = ° - u gT sin 60 = - 10 3 2 10 2 3 = - æ è ç ö ø ÷ 10 3 4 2 3 = - 5 3 m/s v v v x y net = + 2 2 = + - æ è ç ö ø ÷ 5 5 3 2 2 = + 5 1 1 3 = 10 3 m/s 2. Component of velocity perpendicular to plane =v net cosb = ´ ° 10 3 30 cos = ´ 10 3 3 2 =5 m/s 3. Let the particle collide at time t. x u t 1 = ( cos ) q and x v t 2 = \ d x x = - 2 1 = + ( cos ) v u t q = + ° [ cos ] 10 10 2 45 t = 20 t Using equation, s ut at = + 1 2 2 For vertical motion of particle 1 : h u t g t - = + - 10 1 2 2 ( sin ) ( ) q i.e., h u t gt = + - 10 1 2 2 ( sin ) q â€¦(i) or h t gt - + - 10 10 1 2 2 For the vertical motion of particle 2 : 20 1 2 2 - = h gt i.e., h gt = - 20 1 2 2 â€¦(ii) Comparing Eqs. (i) and (ii), 10 10 1 2 20 1 2 2 2 + - = - t g t g t Þ t = 1 s \ d = 20 m 4. u = 10 m/s v = 5 2 m/s Projectile Motion 53 h v u q x 2 x 1 d 10 m 2 1 q f B A d v u a b u a = 60° b = 30° 2 u = 10 m/s 2 g = 10 m/s v x v net v y Page 4 Introductory Exercise 4.1 1. A particle projected at any angle with horizontal will always move in a plane and thus projectile motion is a 2-dimensional motion. The statement is thus false. 2. At high speed the projectile may go to a place where acceleration due to gravity has some different value and as such the motion may not be uniform accelerated. The statement is thus true. 3. See article 4.1. 4. u = 40 2 m/s, q = ° 45 As horizontal acceleration would be zero. v u u x x = = = cos q 40 m/s s u t u t x x = = = ( cos ) q 80 m A : position of particle at time = 0. B : position of particle at time = t. As vertical acceleration would be - g v u gt y y = - = = u gt sinq = - 40 20 =20 m/s s u t gt y y = × - 1 2 2 = - ´ ´ = 80 1 2 10 2 60 2 m \ v v v x y = + 2 2 = + 40 20 2 2 = 20 5 m/s tan f = = v v y x 20 40 i.e., f = æ è ç ö ø ÷ - tan 1 1 2 s s s x y = + 2 2 = + 80 60 2 2 = 100 m tan a = s s y x = 60 80 i.e., a = æ è ç ö ø ÷ - tan 1 3 4 5. s u t g t y y = + - 1 2 2 ( ) or s u t gt y = - ( sin ) q 1 2 2 or 15 20 1 2 2 = - t gt or t t 2 4 3 0 = + = i.e., t = 1 s and 3 s 6. See figure to the answer to question no. 4. u = 40 m/s, q = ° 60 \ u x = ° = 40 60 20 cos m/s Projectile Motion 4 s y g f v v y v x u x s x A a u u y B Thus, v u x x = = 20 m/s As f = ° 45 tan f = = v v y x 1 \ y v y x = = 20 m/s Thus, v v v v x y x = + = 2 2 2 = 20 2 m/s Before reaching highest point v u g t y y = + - ( ) \ 20 40 60 10 = ° - sin t or 2 2 3 = - t Þ t = - 2 3 1 ( ) s After attaining highest point - = - 20 40 3 10t i.e., - = - 2 2 3 t or t = + 2 3 1 ( ) s 7. Average velocity = Range Time of flight = u g u g 2 2 sin 2a a / / sin = ´ 2 2 2 u g g u sin cos sin a a a = ucos a 8. Change in velocity = - - + ( sin ) ( sin ) u u a a = -2usina =2usina (downward) 9. Formulae for R T , and H max will be same if the projection point and the point where the particle lands are same and lie on a horizontal line. 10. r i j ® = + - [ ( ) ] ^ ^ 3 4 5 2 t t t m y-coordinate will be zero when 4 5 0 2 t t - = i.e., t = 0 s, 4 5 s t = 0 belongs to the initial point of projection of the particle. \ r i ® = 0 ^ m i.e., x = 0 m At t = 0.8 s, r i ® = 2.4 ^ m i.e., x = 2.4 m 11. v u x x = = 10 m/s =10 m/s v u x x = = 10 m/s v v x = f cos Þ cosf= v v x = 10 v = 10 10 [as, v = 20 2 (given)] =1 i.e., f = ° 0 \ Speed will be half of its initial value at the highest point where f = ° 0 . Thus, t u g = sin q = ° = 20 60 10 3 sin s 52 | Mechanics-1 a a u â€“ u sin a u sin a t = T u cos a + u sin a u 60° u = 20 m/s v x v f u = 20 cos 60° x Introductory Exercise 4.2 1. Time of flight T u g = - 2 sin ( ) cos a b b = ´ ° - ° ° ( sin( ) cos 2 10 60 30 10 30 = 2 3 s Using, v u at = + v u u x x = = ° cos 60 = ´ = 10 1 2 5 m/s v u g T y y = + - ( ) = ° - u gT sin 60 = - 10 3 2 10 2 3 = - æ è ç ö ø ÷ 10 3 4 2 3 = - 5 3 m/s v v v x y net = + 2 2 = + - æ è ç ö ø ÷ 5 5 3 2 2 = + 5 1 1 3 = 10 3 m/s 2. Component of velocity perpendicular to plane =v net cosb = ´ ° 10 3 30 cos = ´ 10 3 3 2 =5 m/s 3. Let the particle collide at time t. x u t 1 = ( cos ) q and x v t 2 = \ d x x = - 2 1 = + ( cos ) v u t q = + ° [ cos ] 10 10 2 45 t = 20 t Using equation, s ut at = + 1 2 2 For vertical motion of particle 1 : h u t g t - = + - 10 1 2 2 ( sin ) ( ) q i.e., h u t gt = + - 10 1 2 2 ( sin ) q â€¦(i) or h t gt - + - 10 10 1 2 2 For the vertical motion of particle 2 : 20 1 2 2 - = h gt i.e., h gt = - 20 1 2 2 â€¦(ii) Comparing Eqs. (i) and (ii), 10 10 1 2 20 1 2 2 2 + - = - t g t g t Þ t = 1 s \ d = 20 m 4. u = 10 m/s v = 5 2 m/s Projectile Motion 53 h v u q x 2 x 1 d 10 m 2 1 q f B A d v u a b u a = 60° b = 30° 2 u = 10 m/s 2 g = 10 m/s v x v net v y q = ° 30 f = ° 45 d = 15 m Let the particles meet (or are in the same vertical time t). \ d u t v t = + f ( cos ) ( cos ) q Þ 15 10 30 5 2 45 = ° + ° ( cos cos ) t or 15 5 3 5 = + ( ) t or t = + 3 3 1 s = 1.009 s Now, let us find time of flight of A and B T u g A = 2 sin q = 1 s As T t A < , particle A will touch ground before the expected time t of collision. \ Ans : NO. 5. For range to be maximum a p b = + 4 2 = + p p 4 6 2 / = p 3 = ° 60 6. At point A velocity ( ) v ® of the particle will be parallel to the inclined plane. u = 40 m/s a = ° 60 b = ° 30 g = 10 m/s 2 \ f = b v u u x x = = cos a v v v x = f = cos cos b or u v cos cos a b = Þ v u = = ° ° cos cos cos cos a b 40 60 30 = æ è ç ö ø ÷ 40 1 2 3 2 = 40 3 m/s 7. (a) At time t, vertical displacement of A = Vertical displacement of B ( sin ) v t gt v t gt A B q - = - 1 2 1 2 2 2 i.e., v v A B sin q = sin q = v v B A = = 10 20 1 2 \ q = ° 30 (b) x v t A = ( cos ) q = ° ´ ( cos ) 20 30 1 2 = 5 3 m 54 | Mechanics-1 a b = 30° = u p 6 a b u A f A v time t v = 10 m/s B 2 a = â€“ 10 m/s v = 20 m/s A A B x Page 5 Introductory Exercise 4.1 1. A particle projected at any angle with horizontal will always move in a plane and thus projectile motion is a 2-dimensional motion. The statement is thus false. 2. At high speed the projectile may go to a place where acceleration due to gravity has some different value and as such the motion may not be uniform accelerated. The statement is thus true. 3. See article 4.1. 4. u = 40 2 m/s, q = ° 45 As horizontal acceleration would be zero. v u u x x = = = cos q 40 m/s s u t u t x x = = = ( cos ) q 80 m A : position of particle at time = 0. B : position of particle at time = t. As vertical acceleration would be - g v u gt y y = - = = u gt sinq = - 40 20 =20 m/s s u t gt y y = × - 1 2 2 = - ´ ´ = 80 1 2 10 2 60 2 m \ v v v x y = + 2 2 = + 40 20 2 2 = 20 5 m/s tan f = = v v y x 20 40 i.e., f = æ è ç ö ø ÷ - tan 1 1 2 s s s x y = + 2 2 = + 80 60 2 2 = 100 m tan a = s s y x = 60 80 i.e., a = æ è ç ö ø ÷ - tan 1 3 4 5. s u t g t y y = + - 1 2 2 ( ) or s u t gt y = - ( sin ) q 1 2 2 or 15 20 1 2 2 = - t gt or t t 2 4 3 0 = + = i.e., t = 1 s and 3 s 6. See figure to the answer to question no. 4. u = 40 m/s, q = ° 60 \ u x = ° = 40 60 20 cos m/s Projectile Motion 4 s y g f v v y v x u x s x A a u u y B Thus, v u x x = = 20 m/s As f = ° 45 tan f = = v v y x 1 \ y v y x = = 20 m/s Thus, v v v v x y x = + = 2 2 2 = 20 2 m/s Before reaching highest point v u g t y y = + - ( ) \ 20 40 60 10 = ° - sin t or 2 2 3 = - t Þ t = - 2 3 1 ( ) s After attaining highest point - = - 20 40 3 10t i.e., - = - 2 2 3 t or t = + 2 3 1 ( ) s 7. Average velocity = Range Time of flight = u g u g 2 2 sin 2a a / / sin = ´ 2 2 2 u g g u sin cos sin a a a = ucos a 8. Change in velocity = - - + ( sin ) ( sin ) u u a a = -2usina =2usina (downward) 9. Formulae for R T , and H max will be same if the projection point and the point where the particle lands are same and lie on a horizontal line. 10. r i j ® = + - [ ( ) ] ^ ^ 3 4 5 2 t t t m y-coordinate will be zero when 4 5 0 2 t t - = i.e., t = 0 s, 4 5 s t = 0 belongs to the initial point of projection of the particle. \ r i ® = 0 ^ m i.e., x = 0 m At t = 0.8 s, r i ® = 2.4 ^ m i.e., x = 2.4 m 11. v u x x = = 10 m/s =10 m/s v u x x = = 10 m/s v v x = f cos Þ cosf= v v x = 10 v = 10 10 [as, v = 20 2 (given)] =1 i.e., f = ° 0 \ Speed will be half of its initial value at the highest point where f = ° 0 . Thus, t u g = sin q = ° = 20 60 10 3 sin s 52 | Mechanics-1 a a u â€“ u sin a u sin a t = T u cos a + u sin a u 60° u = 20 m/s v x v f u = 20 cos 60° x Introductory Exercise 4.2 1. Time of flight T u g = - 2 sin ( ) cos a b b = ´ ° - ° ° ( sin( ) cos 2 10 60 30 10 30 = 2 3 s Using, v u at = + v u u x x = = ° cos 60 = ´ = 10 1 2 5 m/s v u g T y y = + - ( ) = ° - u gT sin 60 = - 10 3 2 10 2 3 = - æ è ç ö ø ÷ 10 3 4 2 3 = - 5 3 m/s v v v x y net = + 2 2 = + - æ è ç ö ø ÷ 5 5 3 2 2 = + 5 1 1 3 = 10 3 m/s 2. Component of velocity perpendicular to plane =v net cosb = ´ ° 10 3 30 cos = ´ 10 3 3 2 =5 m/s 3. Let the particle collide at time t. x u t 1 = ( cos ) q and x v t 2 = \ d x x = - 2 1 = + ( cos ) v u t q = + ° [ cos ] 10 10 2 45 t = 20 t Using equation, s ut at = + 1 2 2 For vertical motion of particle 1 : h u t g t - = + - 10 1 2 2 ( sin ) ( ) q i.e., h u t gt = + - 10 1 2 2 ( sin ) q â€¦(i) or h t gt - + - 10 10 1 2 2 For the vertical motion of particle 2 : 20 1 2 2 - = h gt i.e., h gt = - 20 1 2 2 â€¦(ii) Comparing Eqs. (i) and (ii), 10 10 1 2 20 1 2 2 2 + - = - t g t g t Þ t = 1 s \ d = 20 m 4. u = 10 m/s v = 5 2 m/s Projectile Motion 53 h v u q x 2 x 1 d 10 m 2 1 q f B A d v u a b u a = 60° b = 30° 2 u = 10 m/s 2 g = 10 m/s v x v net v y q = ° 30 f = ° 45 d = 15 m Let the particles meet (or are in the same vertical time t). \ d u t v t = + f ( cos ) ( cos ) q Þ 15 10 30 5 2 45 = ° + ° ( cos cos ) t or 15 5 3 5 = + ( ) t or t = + 3 3 1 s = 1.009 s Now, let us find time of flight of A and B T u g A = 2 sin q = 1 s As T t A < , particle A will touch ground before the expected time t of collision. \ Ans : NO. 5. For range to be maximum a p b = + 4 2 = + p p 4 6 2 / = p 3 = ° 60 6. At point A velocity ( ) v ® of the particle will be parallel to the inclined plane. u = 40 m/s a = ° 60 b = ° 30 g = 10 m/s 2 \ f = b v u u x x = = cos a v v v x = f = cos cos b or u v cos cos a b = Þ v u = = ° ° cos cos cos cos a b 40 60 30 = æ è ç ö ø ÷ 40 1 2 3 2 = 40 3 m/s 7. (a) At time t, vertical displacement of A = Vertical displacement of B ( sin ) v t gt v t gt A B q - = - 1 2 1 2 2 2 i.e., v v A B sin q = sin q = v v B A = = 10 20 1 2 \ q = ° 30 (b) x v t A = ( cos ) q = ° ´ ( cos ) 20 30 1 2 = 5 3 m 54 | Mechanics-1 a b = 30° = u p 6 a b u A f A v time t v = 10 m/s B 2 a = â€“ 10 m/s v = 20 m/s A A B x AIEEE Cor ner Subjective Questions (Level 1) 1. (a) R u g = = ° 2 2 2 20 2 90 10 sin ( ) sin q = 80 m H u g = = ° 2 2 2 2 2 20 2 45 10 sin ( ) sin q = 40 m T u g = = ° 2 2 20 2 45 10 sin ( ) sin q = 4 s (b) u i j ® = + ( ) ^ ^ 20 20 m/s and a j ® = - 10 ^ m/s 2 - = + ® ® ® v u a t = + + - ( ) ( ) ^ ^ ^ 20 20 10 1 i j j (at t = 1 s) = - ( ) ^ ^ 20 10 i j m/s (c) Time of flight T u g = 2 sin q = ´ ´ æ è ç ö ø ÷ 2 20 2 1 2 10 = 4 s (c) \ Velocity of particle at the time of collision with ground. = + + - ( ) ( ) ^ ^ ^ 20 20 10 4 i j j = - ( ) ^ ^ 20 20 i j m/s 2. (a) s ut at = + 1 2 2 \ ( ) ( ) ( ) - = + + - 40 20 1 2 10 2 T T or 5 20 40 0 2 T T - - = or T T 2 4 8 0 - - = \ T = - - + - - - ( ) ( ) ( )( ) ( ) 4 4 4 1 8 2 1 2 Leaving - ive sign which is not positive. = + 4 48 2 = + 4 4 3 2 = + ( ) 2 2 3 s =5.46 s R = ´ = 20 5.46 109.2 m (b) s ut at = + 1 2 2 ( ) ( ) - = ´ + - 40 1 2 10 2 0 T T Þ T = 2 2 s = 2.83 s R = ´ 2.83 20 = 56.6 m (c) s ut at = + 1 2 2 ( ) ( ) ( ) - = - + - 40 20 1 2 10 2 T T or 5 20 40 0 2 T T + - = or T T 2 4 8 0 + - = \ T = - + - - 4 4 4 1 8 2 1 2 ( ) ( ) ( ) ( ) = - + 4 4 3 2 = - + 2 2 3 = 1.46 s Projectile Motion 55 R â€“ 40 m + T 2 a = â€“ 10 m/s 45° 20Ö2 m/s 20 m/s R â€“ 40 m + T 20 m/s 2 a = â€“ 10 m/s R T â€“ 40 m 20 Ö2 m/sRead More

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