Projectile Motion, Problems to calculate time of flight, Physics Assignment Help

Assignment Help: >> Motion in two Dimension >> Projectile Motion, Problems to calculate time of flight, Physics

 

Let P(x, y) present the location of a projectile after a time t from the time of projection.

2487_Projectile Motion.png

At the point O, horizontal part of velocity = u cosΘ and vertical part of velocity = u sinΘ

Acceleration due to gravity performs vertically downwards. Hence vertical element of velocity modifies. While horizontal element of velocity remains constant (u cosΘ ) in the whole motion.

 

Time of flight: It is the time taken by the body from the instant it is realized till it strikes a position on the same horizontal plane as the point of projection.

To calculate time of flight, we can find the time when vertical displacement is zero i.e.

0 = u sinΘ t - 1/2gt2 327_linear motion.png (u sinΘ - 1/2gt) t = 0   327_linear motion.png    t = 0     or         t = 2usinΘ/g

Thus ,  Time of flight T =2usinΘ/g

Range: During this time horizontal element of velocity has taken the particle through a distance x horizontally

Where,            x = u cosΘ * T = u cosΘ * (2usinΘ/g) = u2sin2Θ/g

This distance, we call as range on horizontal plane , R=u2sin2Θ/g   

            Rmax = u2/g      for Θ = 45°

For a given velocity same range can be obtained for an angle Θ and angle (90° - Θ) i.e.

 R = u2sin2(90o-Θ)/g = u2sin2(180o-Θ)/g = u2sin2Θ/g     

To calculate maximum height, the vertical component of velocity equals to zero, when the particle is at the highest point from the ground. At that time particle has only horizontal component of velocity i.e. = u cosΘ

            0 = (u sinΘ)2 - 2gHmax

            Hmax= u2sin2Θ/2g

The motion of projectile can be analyzed through vector notation also for e.g. velocity of projectile (v) at any time t can be written as

692_Projectile Motion1.png

Hence, the magnitude of velocity (speed) v at any time is given as ,

         453_Projectile Motion2.png  (The path of a projectile is called its trajectory)

Trajectory of projectile

y = u sinΘ t - 1/2gt2                 . . . (1)      (vertical displacement at any time t)

x = u cosΘ t                             . . . (2)      (Horizontal displacement at any time t)

Using t =x/ucosΘ

We get, y=xtanΘ-gx2/2ucos2Θ Which suggest that trajectory of projectile is parabola

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