# Projectile Motion

### Oblique projection on a horizontal surface:

Let a particle (body) be projected with certain velocity $\vec{v_o}$  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

$\vec{r} = x \hat{i} + y \hat{j}$   ….(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 (v0)x = v0cosθ0 .

⇒ vx = v0cosθ0       ……(ii)

⇒ The horizontal distance covered during time t is given as ,  x = (v0x)t

⇒ x = (v0cosθ0)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/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 ,

On putting (v0)y = v0sinθ0 we obtain

vy = v0sinθ0 – gt     …..(iv)

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

vy2 = (voy)2 – 2gy

On putting (voy) = vo sinθ we obtain

$\displaystyle v_y = \sqrt{(v_0 sin\theta_0)^2 – 2 g y }$ ….(v)

Now the vertical displacement y is given as

$\displaystyle y = v_{oy} t – \frac{1}{2}g t^2$

⇒ $\displaystyle y = v_o sin\theta_0 t – \frac{1}{2}g t^2$ ….(vi)

$\displaystyle \vec{r} = x \hat{i} + y \hat{j}$

$\displaystyle \vec{r} = (v_o cos\theta_0 )t \hat{i} + (v_o sin\theta_0 t – \frac{1}{2}g t^2) \hat{j}$

$\displaystyle r = \sqrt{(v_o cos\theta_0 t)^2 + (v_o sin\theta_0 t – \frac{1}{2}g t^2)^2 }$

$\displaystyle tan\theta = \frac{y}{x}$

$\displaystyle tan\theta = \frac{(v_o sin\theta_0 t – \frac{1}{2}g t^2)}{(v_o cos\theta_0 )t}$

Solved Example : 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

$\displaystyle \vec{r} = x \hat{i} + y \hat{j}$

where x = (v0 cos θ0)t

x = (10 cos30°)(1/2)

= 4.33 m.

$\displaystyle y = v_{oy} t – \frac{1}{2}g t^2$

⇒ $\displaystyle y = v_o sin\theta_0 t – \frac{1}{2}g t^2$

$\displaystyle y = 10 sin30 (\frac{1}{2}) – \frac{1}{2}(10)(\frac{1}{2})^2$

y = 1.25 m

$\displaystyle \vec{r} = 4.33\hat{i} + 1.25 \hat{j}$

Exercise : Referring to the previous illustration find the position of the particle as t = v0/g

### Average Velocity :

Therefore the average velocity of the particle during time t can be found as

$\displaystyle \vec{v_{av}} = \frac{\vec{\Delta r}}{\Delta t}$

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,

$\displaystyle \vec{\Delta r} = \vec{r}$ , Putting Δt = t we obtain,

$\displaystyle |\vec{v_{av}}| = \frac{|\vec{r}|}{t}$

### How to find 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

$\displaystyle \vec{v} = v_x \hat{i} + v_y \hat{j}$

Putting the values of vx and vy using above Equation , we obtain

$\displaystyle \vec{v} = v_0 cos\theta_0 \hat{i} + ( v_0 sin\theta_0 – g t) \hat{j}$

$\displaystyle v = \sqrt{ (v_0 cos\theta_0)^2 + ( v_0 sin\theta_0 – g t)^2}$

$\displaystyle tan\theta = \frac{v_y}{v_x}$

$\displaystyle tan\theta = \frac{( v_0 sin\theta_0 – g t)}{(v_0 cos\theta_0)}$