Friday, July 5, 2013


Projectile Motion Equation is used to find the distance, velocity and time taken in the projectile motion. The Curved path along which the projectile travels is what is known as trajectory. Projectile Motion is the free fall motion of any body in a horizontal path with constant velocity. 

Where Vx is the velocity along x-axis,
       Vxo is the initial velocity along x-axis,
          Vy is the velocity along y-axis,
          Vyo is the initial velocity along y-axis.
  g is the acceleration due to gravity and
         t is the time taken.

 Where Vo is the initial Velocity
 sin θ is the component along y-axis,
 cos θ is the component along x-axis.

Solved Examples 

Question 1:  

A body is projected with a velocity of 20 ms-1 at 50o to the horizontal. Find
 (i) Maximum height reached
(ii) Time of flight and
(iii) Range of the projectile.
Initial Velocity Vo = 20 ms-1,
θ = 50o

Time of flight, t = 2V0sinθg
                           = 2×20×sin5009.8
                           = 3.126 s.

Maximum Height reached, H = V2osin2θ2g
                                                    = (20)2sin25022×9.8
                                                    = 11.97 m.

Horizontal Range R = v20sin2θg
= 202sin100o9.8
                                    = 40.196 m.

Question 2:
John is on top of the building and jack is down. If john throws a ball at an angle of 60o and with initial velocity 20 m/s. At what height will the ball reach after 2 s?
Solution: Given: Vyo = 20 m/s,
           Δ t = 2s,
The Vertical velocity in y direction is given by Vy = Vyo sin 60o
                                                                                     = 20 × 32
                                                                                     = 17.32 m/s.
 Vertical distance, y = Vyo t - 12 gt2
                                   = 20 × 2 - 0.5 × 9.8 × 4
                                   = 20.4 m.

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