{"id":6811,"date":"2022-07-12T07:17:24","date_gmt":"2022-07-12T07:17:24","guid":{"rendered":"https:\/\/physics.educour.in\/isc-physics\/?p=6811"},"modified":"2026-06-05T09:33:54","modified_gmt":"2026-06-05T09:33:54","slug":"projectile-motion","status":"publish","type":"post","link":"https:\/\/physics.educour.in\/isc-physics\/projectile-motion\/","title":{"rendered":"Projectile Motion"},"content":{"rendered":"\t\t
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Projectile Motion<\/span><\/h1>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0Any object that is projected with some initial velocity in any direction other than vetical and that subsequently allowed to move freely under the gravitational force acting on it, is called a Projectile<\/strong><\/span>. A projectile may be a football, a cricket ball, shell of a gun or any other object.<\/span><\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t

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Note: <\/span><\/strong>
(i) The motion of projectile is known as projectile motion.<\/span>
(ii) It is an example of two dimensional motion with constant acceleration.<\/span>
(iii) The path followed by a particle (here projectile) during its motion is called its Trajectory. <\/span>
(iv) In this chapter we shall consider only trajectories that are of sufficiently short range so that the gravitational force can be considered constant in both magnitude and direction. <\/span><\/em><\/span>
(v) All effects of air resistance will be ignored; thus our results are precise only for motion in a vacuum on flat, non rotating Earth.<\/span><\/em><\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t

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PROJETILE MOTION AS TWO SIMULTANIOUS MOTIONS (The Two Components of Projectile Motion)<\/strong><\/span><\/p>

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0Projectile 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<\/span><\/p>

Along the x-axis:<\/strong> uniform velocity, responsible for the horizontal (forward) motion of the particle.<\/span>
Along y-axis:<\/strong> uniform acceleration, responsible for the vertical (downwards) motion of the particle.<\/span><\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t

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\"\"<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t

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\u00a0 \u00a0 \u00a0 \u00a0 =<\/h2>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t
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\"\"<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Parabolic path = vertical motion + horizontal motion.<\/span><\/h4>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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Oblique Projectile<\/span><\/h2>

(Projectile thrown at an angle with horizontal)<\/span><\/h4>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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(i) Consider a projectile thrown with a velocity u making an angle q0<\/sub><\/span> with the horizontal.<\/span>
(ii) Initial velocity u is 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.<\/span>
ux<\/sub> = u cos q0<\/sub><\/span><\/span>
uy<\/sub> = u sin q0<\/sub><\/span><\/span><\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t

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(iii) Again this projectile motion can be considered as the combination of horizontal and vertical motion. Therefore,<\/span><\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t

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\"\"<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Horizontal direction (x- axis)<\/span><\/strong><\/h5>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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(a) Initial velocity ux<\/sub> = u cosq0<\/sub><\/span>\u00a0<\/span><\/strong><\/span>
(b) Acceleration ax<\/sub> = 0\u00a0<\/span><\/strong><\/span>
(c) Velocity after time t,\u00a0<\/span><\/strong><\/span><\/p>

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 vx<\/sub> = u cos q0<\/sub><\/span>\u00a0<\/span><\/strong><\/span><\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t

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\u00a0Vertical direction (y-axis)<\/span><\/strong><\/h5>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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(a)Initial velocity uy<\/sub> = u sin q0<\/sub><\/span><\/span><\/strong>
(b) \u00a0Acceleration ay<\/sub> = g<\/span><\/strong>
(c) Velocity after time t,<\/span><\/strong><\/p>

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0vy<\/sub> = u sin q0<\/sub><\/span> \u2013 gt<\/span><\/strong><\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Time of Flight:<\/span>\u00a0<\/strong><\/h3>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\u00a0Consider the whole motion, from point O to point B:<\/span><\/p>

Since the displacement along vertical direction does not occur. So, for t = T, Sy<\/sub> = 0<\/span><\/p>

\u00a0 \u00a0 \u00a0 \u00a0uy<\/sub>.t + \u00bd gt2<\/sup> = Sy<\/sub> = 0<\/span><\/p>

Or\u00a0 \u00a0 \u00a0(u sin q0<\/sub><\/span>)T \u2013 \u00bd gT2<\/sup> = 0<\/span><\/p>

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0\"\"\u00a0 \u00a0…………. [1]
<\/span><\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t

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Maximum Height: <\/strong><\/span><\/h3>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0Consider the motion from lowest point O and the highest point A. At the highest point, vertical component of velocity vy<\/sub> =0 (As t<\/span>he tangent to the path is horizontal and hence velocity vector is horizontal, i.e there is no vertical component of velocity)<\/span><\/p>

Using IIIrd<\/sup> eqn. of motion i.e.<\/span><\/p>

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 vy<\/sub>2<\/sup> = uy<\/sub>2<\/sup> + 2ay<\/sub> sy<\/sub><\/span><\/p>

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 we have for vertical direction<\/span><\/p>

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a00 = u2<\/sup>sin2<\/sup>q0<\/sub><\/span> – 2gH<\/span><\/p>

Or\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0<\/span>\"\"\u00a0 \u00a0 …………….[2]<\/span><\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t

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Horizontal range: <\/strong><\/span><\/h3>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0The horizontal distance covered by the projectile during its time of flight is known as the horizontal range of the projectile\u00a0<\/span><\/p>

Consider the whole motion, from point O to point B:\u00a0<\/span><\/p>

Along x-axis,\u00a0 Sx<\/sub> = R and t =T<\/span><\/p>

Using the formula\u00a0 \u00a0<\/span><\/p>

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0Sx<\/sub> = ux<\/sub>.t + \u00bd ax<\/sub> t2<\/sup><\/span><\/p>

Or\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0R = ux<\/sub> . T<\/span><\/p>

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<\/span><\/p>

Or\u00a0 \u00a0 \u00a0\"\"\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 …………………..[3]<\/span><\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t

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Note:\u00a0<\/span><\/strong><\/span><\/span><\/p>

  • Equations [1], [2] and [3] are valid only for complete flight, that is when proejctile lands at same horizontal level from which it has been projected.<\/span><\/li>
  • Vertical component of velocity is zero when particle moves horizontally, i.e., at the highest point of trajectory.<\/span><\/li>
  • Vertical component of velocity is positive when particle is moving up and vertical component of velocity is negative when particle is coming down if vertical upwards direction is taken as positive. Any direction upward or downward can be taken as positive and if downward direction is taken as positive then vertical component of velocity coming down is positive.<\/span><\/li><\/ul>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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    Projectile Motion [Part 1]<\/span><\/h4>
    (In Hindi + English Mix)<\/span><\/h5>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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