**Oblique projection on a horizontal surface:**

**Let a particle (body) be projected with certain velocity v _{0}^{→} at an angle θ_{0} to the horizontal.**

**The horizontal component of its velocity , (v _{ox}) = v_{0}cosθ_{0}**

**The vertical component of its velocity ,( v _{oy}) = v_{0}sinθ_{0}**

**The particle moves simultaneously in both horizontal and vertical directions under earths gravitational field (no other external forces like wind drag are small and therefore their effect neglected.**

**Change in Position Vector (Displacement) :**

**Let the particle acquire a position P having the coordinates (x , y) just after time t from the instant of projection . The corresponding position vector of the particle at time t is as shown in the Figure**

** ….(i)**

**Since no external force acts upon the particle horizontally, its horizontal acceleration is zero, that means, the particle moves horizontally with constant velocity of magnitude (v _{0})x = v_{0}cosθ_{0} .**

**⇒ v _{x} = v_{0}cosθ_{0 } ……(ii)**

**⇒ The horizontal distance covered during time t is given as , x = (v _{0x})t**

**⇒ x = (v _{0}cosθ_{0})t ….(iii)**

**If we consider the vertical motion of the particle, the external force acting on the particle is gravitational force mg.**

**Consequently the particle accelerates downwards (towards the centre of earth) with an acceleration of magnitude g = 9.8 m/sec ^{2}**

**In the other words we can say that the particle decelerates upward with g = 9.8 m/sec ^{2}**

**Consequently the vertical velocity of the particle at time t is given as , v _{y} = (v_{0})_{y} – gt ,**

**On putting (v _{0})_{y} = v_{0}sinθ_{0} we obtain**

**v _{y} = v_{0}sinθ_{0} – gt …..(iv)**

**The vertical velocity of the particle at any height y is given as**

**v _{y}^{2} = (v_{oy})^{2} – 2gy , putting (v_{oy}) = v_{o} sinθ we obtain**

** ….(v)**

**Now the vertical displacement y is given as**

**⇒ ….(vi)**

**Illustration : A boy throws a stone with an speed V _{0} = 10 m/sec at an angle θ_{0} = 30° to the horizontal. Find the position of the stone w.r.t. the point of projection just after a time t = 1/2 sec.**

**Solution: **

**The position of the stone is given by**

**where x = (v _{0} cos θ_{0})t**

**x = (10 cos30°)(1/2)**

**= 4.33 m.**

**⇒ **

**y = 1.25 m**

**Exercise : Referring to the previous illustration find the position of the particle as t = v _{0}/g**

__Average Velocity :__

__Average Velocity :__**Therefore the average velocity of the particle during time t can be found as**

**We have assumed the point of projection as the origin of the coordinate system.**

**That means, the initial position vector of the particle has a magnitude equal to zero,**

** , Putting Δt = t we obtain,**

__Instantaneous Velocity__

__Instantaneous Velocity__**The velocity of the particle at time t is equal to the vector sum of the velocity components along x and y axis**

**Putting the values of v _{x} and v_{y} using above Equation , we obtain**