Projectile Motion , Time of Flight , Maximum Height , Horizontal Range

How to find Equation of Path of Projectile ?

Let a Particle be projected with velocity v0 and after time t it acquire a position P(x , y)

x = v0cosθ0 t

$ \displaystyle t = \frac{x}{v_0 cos\theta_0} $

$ \displaystyle y = v_{0y} t – \frac{1}{2}g t^2$

$ \displaystyle y = v_0 sin\theta_0 t – \frac{1}{2}g t^2 $

$ \displaystyle y = v_0 sin\theta_0 (\frac{x}{v_0 cos\theta_0}) – \frac{1}{2}g (\frac{x}{v_0 cos\theta_0})^2 $

$ \displaystyle y = x tan\theta_0 – \frac{g x^2}{2 v_0^2 cos^2\theta_0}$

This is the equation of a parabola. Therefore the path of the particle is parabolic when the particle passes the level of projection.

How to Calculate Time of Flight (T) ?

The horizontal distance covered is known as its range R and the time of motion is known as time of flight T.

$ \displaystyle y = v_0 sin\theta_0 t – \frac{1}{2}g t^2 $

When t = T , y = 0

$ \displaystyle 0 = (v_0 sin\theta_0 )T – \frac{1}{2}g T^2 $

$ \displaystyle (v_0 sin\theta_0 )T = \frac{1}{2}g T^2 $

$ \displaystyle T = \frac{2v_0 sin\theta_0 }{g } $

How to Calculate Maximum Height (H) ?

$ \displaystyle {v_y}^2 = v_{0y}^2 – 2 g y $

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

When the particle is at the highest position, vy = 0 and y = H ,

$ \displaystyle 0 = ({v_0 sin\theta_0})^2 – 2 g H $

$ \displaystyle H = \frac{v_0^2 sin^2\theta_0}{2 g } $

How to Calculate Horizontal Range (R) ?

The Equation of trajectory is ,

$ \displaystyle y = x tan\theta_0 – \frac{g x^2}{2 v_0^2 cos^2\theta_0}$

When x = R , y = 0 ,

$ \displaystyle 0 = R tan\theta_0 – \frac{g R^2}{2 v_0^2 cos^2\theta_0}$

$ \displaystyle R tan\theta_0 = \frac{g R^2}{2 v_0^2 cos^2\theta_0}$

$ \displaystyle R = \frac{2v_0^2 sin\theta_0 cos\theta_0}{g} $

$ \displaystyle R = \frac{v_0^2 sin 2\theta_0}{g} $

⇒ R is maximum when Sin2θo is maximum

Sin2θo = 1 (maximum)

⇒ 2θo = 90°

θo = 45°

Solved Example : A body is projected up such that its position vector varies with time as
$ \displaystyle \vec{r} = 6t\hat{i}+ (8t-5t^2)\hat{j} $ . Find the

(a) Initial velocity

(b) Time of flight

(c) Horizontal range of the body.

(d)  Maximum height attained by the body.

(e) Equation of trajectory of the body.

Solution: (a) The position of the body at any time t is given as

$ \displaystyle \vec{r} = 6t\hat{i}+ (8t-5t^2)\hat{j} $ .

When t = 0, r = 0. That means the body is projected from the origin of the coordinate system.

Differentiating both sides w.r.t. time ‘ t ‘, we obtain

$ \displaystyle \frac{\vec{dr}}{dt} = 6\hat{i}+(8-10t)\hat{j} $

$ \displaystyle \vec{v} = 6\hat{i} + (8 – 10t)\hat{j} $

Putting t = 0, we obtain the initial velocity (velocity of projection) given as

$ \displaystyle \vec{v_0} = 6\hat{i} + 8\hat{j} $

v0x = 6 , v0y = 8

v0cosθ0 = 6 , v0sinθ0 = 8

$ \displaystyle v_0 = \sqrt{6^2 + 8^2} $

vo = 10 m/sec;

(b) The time of flight $ \displaystyle T = \frac{2v_0 sin\theta_0 }{g } $

$ \displaystyle T = \frac{2v_{0y}}{g } $ , where (v0y) = 8

$ \displaystyle T = \frac{2 \times 8}{10} $

⇒  T = 1.6 sec.

Solved Example : A ball is thrown from ground level so as to just clear a wall 4 metres high at a distance of 4 metres and falls at a distance of 14 metres from the wall. Find the magnitude and direction of the velocity of the ball.

Solution:

let P be a point on the trajectory whose coordinates are (4, 4).

As the ball strikes the ground at a distance 14 metre from the wall, the range is 4 + 14 = 18 metre.

The equation of trajectory is

$ \displaystyle y = x tan\theta_0 – \frac{g x^2}{2 v_0^2 cos^2\theta_0}$

$ \displaystyle y = x tan\theta_0 [ 1- \frac{g x}{2 v_0^2 cos^2\theta_0 . tan\theta_0} ] $

$ \displaystyle y = x tan\theta_0 [ 1- \frac{x}{2 v_0^2 sin\theta_0 cos\theta_0 /g } ] $

$ \displaystyle y = x tan\theta_0 [ 1 – \frac{x}{R}] $

Here x = 4 , y = 4 and R = 18

$ \displaystyle 4 = 4 tan\theta_0[ 1 – \frac{4}{18}] $

$ \displaystyle 4 = 4 tan\theta_0 ( \frac{7}{9}) $

tanθ = 9/7

$ \displaystyle sin\theta_0 = \frac{9}{\sqrt{130}} \; cos\theta_0 = \frac{7}{\sqrt{130}} $

Again , $ R = \frac{2 v_0^2 sin\theta_0 cos\theta_0}{g} $

On putting the given value ,

$ \displaystyle v_0 = \sqrt{182} m/s $

Exercise : A particle is thrown over a triangle from one end of a horizontal base and grazing the vertex falls on the other end of the base. If α and β be the base angle and θ the angle of projection, prove that tan θ = tan α + tan β

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
→Horizontal Projection from a given height
→ Radius of Curvature at any point on the Path of a Projectile
→ Projectile on an inclined plane

← Back Page | Next Page →