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Projectile motion, , , Projectile, , , , Projectile is an object that is in flight after being thrown or projected., , , , Such a projectile might be a football, a cricket ball, a baseball, etc., , , , eg: 1. An object dropped from aeroplane, , , , 2. A bullet fired from a rifle., , , , 3. Even a jet of water coming out from the side hole of a vessel.

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, , In absence of air resistance, the motion of a projectile is considered as the, combination of the following two independent motions:, , 1., , Motion along horizontal direction with uniform velocity. ( Without any, acceleration), , 2., , Motion along vertical direction under gravity. ( i.e. with uniform acceleration, equal to g), , , , The two motions of a projectile along horizontal and vertical directions are, independent of each other., , , , This is called the Principle of physical independence of motions., , , , [ Note: Gravity acts along vertical and not horizontal; this is the basic funda of, projectile motion], , , , The path followed by the projectile is called trajectory and the projectile’s, trajectory is a parabola.

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Q. A projectile is fired with a velocity u making an angle  with the, horizontal., Show that its trajectory is a parabola., Derive expressions for, 1., , Time of maximum height, , 2., , Time of flight, , 3., , Maximum height, , 4., , Horizontal range, , Solution:

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, , Let us assume that the air resistance has negligible effect on the motion of, the projectile., , , , Suppose a body is projected with initial velocity u, making an angle  with the, horizontal., , , , The velocity u has two rectangular components:, , 1., , The horizontal component u cos, which remains constant throughout the, motion., , 2., , The vertical component usin, which changes with time under the effect of, gravity. ( This component first decreases, becomes zero at the highest point, A, after which it again increases, till the projectile hits the ground., , , , Under the combined effect of above two components, the body follows the, parabolic path OAB, as shown in figure.

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(I), , Equation of trajectory of a projectile., Suppose the body reaches the point P(x,y) after time ‘t’.,  The horizontal distance covered by the body in time t is, , x = Horizontal velocity  time, x = u cos  t, t=, , 𝑥, , u cos, , For vertical motion: u = u sin, a = -g, so the vertical distance covered in time t is given by,, 1, , s = ut + 2 𝑎𝑡 2, , y = u sin  {, y = x tan  -, , 𝑥, 1, 𝑥, } + (−𝑔)(, )2, 2, 𝑢 cos , u cos, 𝑔,  𝑥2, 2, 2, 2𝑢 𝑐𝑜𝑠 , , y = p 𝑥 - q 𝑥2, , where p and q are constants., , this is similar to the equation of parabola y = a 𝑥 + b𝑥 2, Hence the trajectory of a projectile is a parabola.

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(II) Time of maximum height:, Let tm be the time taken by the projectile to reach the maximum height hm., At highest point, vertical component of velocity is zero. i.e. V = 0. a = -g, u = u sin , V = u + at, 0 = u sin  - g tm, g tm = u sin , tm =, , u sin, 𝑔

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(III) Time of flight:, It is the time taken by the projectile from the instant it is projected till it, reaches a point in the horizontal plane of its projection., The body reaches the point B after the time of flight Tf.,  Net vertical displacement covered during the time of flight = 0. i.e. S = 0., s = ut +, , 1, 2, , 𝑎𝑡 2, , 0 = u sin   Tf u sin   Tf =, 2u, , sin, 𝑔, ,  Tf =, , 1, 2, , 1, 2, , 𝑔Tf2, , 𝑔Tf2, , Tf2, =, Tf, , 2 𝑢 sin , , g, ,  Tf = 2 tm,  Time of ascent = Time of descent for symmetrical parabolic path.

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(V) Horizontal range (R), It is the horizontal distance travelled by the projectile during its time of flight., Horizontal range = Horizontal velocity  time of flight, R = u cos , =, , 𝑢2, , 𝑔, , =, , 𝑢2 sin 2, 𝑔, , R=, , 𝑢2 sin 2, 𝑔, , 2 𝑢 sin , , g, , 2 sin  cos, ( since 2 sin cos = sin 2)

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, , Condition for the maximum horizontal range., R=, , 𝑢2 sin 2, 𝑔, , clearly, R will be maximum when sin 2 = 1., Sin 2 = 1 = Sin 90, 2 = 90,  = 45, Thus the horizontal range of a projectile is maximum when it is projected at an angle of 45 with the, horizontal., The maximum horizontal range is given by, 𝑢2 sin 90, 𝑢2 (1), 𝑢2, R=, = 𝑔 = 𝑔, 𝑔, Complementary angles have same Range., eg. (i) 30 and 60, (ii) 15 and 75.

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Problems:, 1. A body is projected at an angle of 15 with velocity 10 m/s. What will be the, horizontal range? ( Take g = 10 m/s2), Solution:, (10)2 sin 2 (15), 𝑢2 sin 2, R=, =, 𝑔, 10, (100) sin 30, =, 10, , 1, 100 ×, 2, =, 10, 50, = = 5m, 10

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2. A ball is thrown at a speed of 20 m/s in the direction of 45 above, horizontal. Find the, , (i) Maximum height, (ii) Range, ( Take g = 10 m/s2), , (iii) Time of flight, , Solution:, (i) hm =, (ii) R =, (iii) Tf =, , 𝑢2 𝑠𝑖𝑛2, , , , =, , 2𝑔, , 𝑢2 sin 2, 𝑔, , 2u sin, 𝑔, , =, , =, , 20, , 1, , ), , 2 𝑠𝑖𝑛2 (45, , =, , 2×10, 20 2 sin 2(45), 10, , 2(20) sin(45), 10, , =, , =, , 400 (2), 2×10, , =, , 200, 20, , (400) sin(90), 10, , (40) ( 12), 10, , =, , 4, 2, , = 10m, , =, =, , 400×1, 10, , 4, 1.41, , = 40m, , = 2.83 𝑠𝑒𝑐𝑠.

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3. A cricket ball is thrown at speed of 28 m/s in a direction 30 above horizontal., Calculate, (a) the maximum height, (b) the time taken by the ball to return to the same level., (c) the horizontal distance from the thrower to the point where the ball returns to, the same level., , Solution:, (a) hm =, , (b) Tf =, (c) R =, , 𝑢2 𝑠𝑖𝑛2, , , , 2𝑔, , 2u sin, , 𝑢2, , 𝑔, sin 2, 𝑔, , =, , =, =, , 28, , ), , 2 𝑠𝑖𝑛2 (30, , 2×9.8, , 2(28) sin (30), 9.8, 28, , 2, , sin 2(30), 9.8, , =, , =, =, , 784, ,  (14), , 19.6, , (56) (12), 9.8, , =, , =, , 196, 19.6, , 28, 9.8, , (784) sin(60), 9.8, , = 10 m, , = 2.85 𝑠𝑒𝑐𝑠., 3, , =, , (784)( 2 ), 9.8, , 1.73, , =, , (784)( 2 ), 9.8, , = 69.28 m

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Uniform CIRCULAR MOTION, , , When an object follows a circular path at a constant speed, the motion of the, object is called uniform circular motion., , , , The word uniform refers to the speed, which is uniform (constant) throughout, the motion.

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 When, , a body is in uniform circular motion, its speed, remains constant but its velocity changes continuously due, to the change in its direction. Hence the motion is, accelerated., ,  A body, , undergoing uniform circular motion is acted upon, by an acceleration which is directed along the radius, towards the centre of the circular path. This acceleration is, called Centripetal acceleration.