# Horizontal Projection Motion , Equation of Trajectory , Time of flight , Horizontal Range

Let a particle be projected with a horizontal velocity V0 , which remains constant along horizontal line due to the absence of any horizontal force. Due to earth’s gravitation the particle acquires vertical velocity Vy at any time t and at any position P(x,y). Let the position vector of this point be $\displaystyle \vec{r}$ .

Vy = (Vy)0 + gt

Since there is no vertical component of V0 initially, (Vy)0 = 0
Vy = g t

And , Vertical displacement is

$\displaystyle y = (V_y)_0 t + \frac{1}{2} g t^2$

Again , $\displaystyle V_y^2 = ( V_y)_0^2 + 2 g y$

$\displaystyle V_y^2 = 0 + 2 g y$

$\displaystyle V_y = \sqrt{2 g y }$

### Displacement :

Now the horizontal displacement x = V0t and the vertical displacement $\large y = \frac{1}{2} g t^2$

Hence , $\large t = \sqrt{\frac{2 y}{g}}$

Since the position vector $\displaystyle \vec{r}= x \hat{i} – y \hat{j}$ , putting the values of x and y, we obtain,

$\displaystyle \vec{r}= V_0 t \hat{i} – \frac{1}{2}g t^2 \hat{j}$

Therefore ,

$\displaystyle r = \sqrt{(V_0 t)^2 + (\frac{1}{2}g t^2)^2}$

$\displaystyle tan\phi = \frac{g t^2/2}{V_0 t}$

$\displaystyle tan\phi = \frac{g t}{2 V_0 }$

### Velocity :

$\displaystyle \vec{V}= V_x \hat{i} – V_y \hat{j}$

$\displaystyle \vec{V}= V_0 \hat{i} – g t \hat{j}$

$\displaystyle V = \sqrt{V_0^2 + (g t)^2}$

$\displaystyle tan\theta = \frac{g t}{V_0}$

Again ,

$\displaystyle \vec{V}= V_x \hat{i} – V_y \hat{j}$

$\displaystyle \vec{V}= V_0 \hat{i} – \sqrt{2 g y} \; \hat{j}$

### Horizontal Range :

If y = H (height of the cliff or height of fall of the projectile), the corresponding horizontal distance (Range R) can be found by putting the values of time of fall $\displaystyle t = \sqrt{\frac{2 H}{g}}$ in the equation x = V0t.

$\displaystyle x = R = V_0 \sqrt{\frac{2 H}{g}}$

On Putting x = R and $\displaystyle t = \sqrt{\frac{2 H}{g}}$

We can find the average velocity and displacement of the particle during the motion when projected horizontally from the top of a cliff of height H with a speed V0.

### Equation of Trajectory

The locus of the path of the particle is given as

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

On putting , t = x/V0

$\displaystyle y = \frac{1}{2}g (x/V_0)^2$

$\displaystyle y = \frac{g x^2}{2 V_0^2}$

It is a parabola.