PROJECTILE MOTION EQUATIONS

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 V_x is the velocity along x-axis,

       V_xo is the initial velocity along x-axis,
          V_y is the velocity along y-axis,
          V_yo is the initial velocity along y-axis.
  g is the acceleration due to gravity and
         t is the time taken.

 

 Where V_o 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 50^o to the horizontal. Find
 (i) Maximum height reached
(ii) Time of flight and
(iii) Range of the projectile.
Solution:

Initial Velocity V_o = 20 ms^-1,
θ = 50^o

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 60^o and with initial velocity 20 m/s. At what height will the ball reach after 2 s?
Solution: Given: V_yo = 20 m/s,
           Δ t = 2s,
The Vertical velocity in y direction is given by V_y = V_yo sin 60^o
                                                                                     = 20 × 3√2
                                                                                     = 17.32 m/s.
 Vertical distance, y = V_yo t - 12 gt²
                                   = 20 × 2 - 0.5 × 9.8 × 4
                                   = 20.4 m.