Projectile Motion

Oblique projection on a horizontal surface
Let a particle (body) be projected with certain velocity v0-> at an angle θ0 to the horizontal.

The horizontal component of its velocity

(vox) = v0cosθ0

The vertical component of its velocity

(voy) = v0sinθ0

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 (v0)x = v0cosθ0 .

⇒ vx = v0cosθ0       ……(2)

⇒ The horizontal distance covered during time t is given as

x = (v0x)t

⇒ x = (v0cosθ0)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/sec2

In the other words we can say that the particle decelerates upward with g = 9.8 m/sec2

Consequently the vertical velocity of the particle at time t is given as

vy = (v0)y – gt ,

putting (v0)y = v0sinθ0 we obtain

vy = v0sinθ0 – gt …..(4)

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

vy2 = (voy)2 – 2gy , putting (voy) = vo sinθ we obtain

…….(5)

Now the vertical displacement y is given as

y = (v0)y t -(1/2) gt2

⇒ y = v0sinθ0t – (1/2)gt2 ……….(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 V0 = 10 m/sec at an angle θ0 = 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 = (v0 cos θ0)t

= (10 cos30°)(1/2)

= 4.33 m.

and y = v0sinθ0 – (1/2)gt2

= (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 = v0/g

Average Velocity Instantaneous Velocity

Therefore the average velocity of the particle during time t can be found as Vav-> = 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 vx and vy using equations (2) and (3) in above Equation , we obtain

θ = tan-1 {tanθ0 – (gt/v0)secθ0}

Substituting the values of vx  and vy by equations (2) and (5) in equation (7) we obtain

Next Page →
← Back Page

Leave a Reply