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