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.

Also Read :

→ Position vector , Instantaneous Velocity
→ Equation of Trajectory , Time of Flight , Max. height & Horizontal Range
→ Angle of Projection for given Ratio of Range & Max. Height
→ Speed & Angle of Projection so that projectile Passes through Two given Points
→ Radius of Curvature at any point on the Path of a Projectile
→ Projectile on an inclined plane

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