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Projectile motionchemistory (4)

Projectile motionchemistory (4)



This project is prepared by GLT Saraswati Bal Mandir school, Nehru Nagar students.

This project is prepared by GLT Saraswati Bal Mandir school, Nehru Nagar students.



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    Projectile motionchemistory (4) Projectile motionchemistory (4) Presentation Transcript

    • NEHRU NAGAR Prepared by:-Siddhant-[XI-A] Saurabh-[X-C]
    • What is Projectile Motion? The motion of a projectile may be thought of as the result of horizontal and vertical components. Both the components act independently
    • Projectile Given AngularProjection
    • Equation of path of projectileSuppose at any time t, the object is at point P (x, y). For motion along horizontal direction, the acceleration ax is zero. The position of the object at any time t is given by,Here, x0 = 0, ux = u cos θ, ax = 0[ Velocity of an object in the horizontal direction is constant]Putting these values in equation (i),⇒ x = ut cos θFor motion along vertical direction, the acceleration ay is −g.The position of the object at any time t along the vertical direction is given by,Here,
    • ∴Putting the value of t from equation (ii),⇒This is an equation of a parabola. Hence, the path of the projectile is a parabola
    • Time of flight Total time for which the object is in flight It is denoted by T. Total time of flight = Time of ascent + Time of descent ∴ T = t + t = 2t [Time of ascent = Time of descent = t] ⇒ At the highest point H, the vertical component of velocity becomes zero. For vertical motion of the object (from 0 to H), ∴
    • Maximum height•Maximum height ‘h’ reached by the projectileFor vertical upward motion from 0 to H, Using the relation we obtain ⇒ That is,
    • Horizontal Range•Horizontal distance covered by the object between its point of projection and the point of hitting theground. It is denoted by R.‘R’ is the distance travelled during time of flight T. ⇒ ⇒ For the maximum horizontal range, sin 2θ = 1 = sin 90° ⇒ 2θ = 90° ⇒ θ = 45° ∴ Maximum horizontal range (Rm) is
    • Examples of Projectile Motion• Launching a Cannon ball
    • Frame of reference: Equations of motion: y X Y v0 Uniform m. Accel. m. ACCL. ax = 0 ay = g = -9.81 gh m/s2 VELC. vx = v0 vy = g t x 0 DSPL. x = v0 t y = h + ½ g t2
    • Important points :In case of projectile motion , the horizontal component of a velocity (u cos θ) , acceleration (g) and mechanical energy remains constant while, speed ,velocity , vertical component of velocity (v sin θ) ,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.
    • Vector diagrams forprojectile motion TIME HORIZONTAL VELOCITY VERTICAL VELOCITY 0s 73.1 m/s, right 19.6 m/s, up 1s 73.1 m/s, right 9.8 m/s, up 2s 73.1 m/s, right 0 m/s 3s 73.1 m/s, right 9.8 m/s, down 4s 73.1 m/s, right 19.6 m/s, down 5s 73.1 m/s, right 29.4 m/s, down 6s 73.1 m/s, right 39.2 m/s, down 7s 73.1 m/s, right 49.0 m/s, down
    •  What two factors would affect projectile motion? ◦ Angle ◦ Initial velocity Initial Velocity Angle
    • Evaluating various infoDetermination of the Time of FlightDetermination of Horizontal Displacement x = vix • tDetermination of the Peak Height y = viy • t + 0.5 • g • t2
    • Applications
    • The use of projectilemotion in sports
    •  Like:- Cricketers know that from which angle they have to hit to get the maximum range.
    • For your co-operation