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

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 } ……(2)

⇒ The horizontal distance covered during time t is given as

x = (v_{0x})t

⇒ x = (v_{0}cosθ_{0})t ….(3)

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 ,

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

v_{y} = v_{0}sinθ_{0} – gt …..(4)

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

…….(5)

Now the vertical displacement y is given as

y = (v_{0})_{y} t -(1/2) gt^{2}

⇒ y = v_{0}sinθ_{0}t – (1/2)gt^{2} ……….(6)

Putting the values of x and y from equation (3) and equation (6) in equation (1) we obtain the position vector at any time t as

**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

= (10 cos30°)(1/2)

= 4.33 m.

and y = v_{0}sinθ_{0} – (1/2)gt^{2}

= (10 sin30°)(1/2) − (1/2)×10×(1/2)^{2}

= 1.25 m

r^{->} = 4.33 i^{^} + 1.25 j^{^}

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

__Average Velocity & ____Instantaneous Velocity__

Therefore the average velocity of the particle during time t can be found as V_{av}^{->} = Δr^{->} /Δt

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 Δr^{->} = r^{->} .Putting Δt = t we obtain,

Substituting the obtained value of | r^{->}| we obtain.

__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

………(7)

Putting the values of v_{x} and v_{y} using equations (2) and (3) in above Equation , we obtain

θ = tan^{-1} {tanθ_{0} – (gt/v_{0})secθ_{0}}

Substituting the values of v_{x} and v_{y} by equations (2) and (5) in equation (7) we obtain