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Equation Of Trajectory

Learn formula for the sum of the roots of a quadratic equation and equation of trajectory, practice graphing quadratic and angle made by trajectory, maximum height of a projectile.

Two Roots of Quadratic Equation

  • In projectile motion, there can be two values of x - coordinate for any specific y - coordinate, as shown in figure.
  • Because the projectile attains a specific height two times during a projectile motion, once in upward motion and once in downward motion.

Values at Maximum Height

  • At maximum height, both the x - coordinates are equal i.e., there is only one value of x at maximum height.                     

  • To understand this phenomenon, consider a man who tries to kill the flying bird with a gun at angle \(\theta\) with horizontal direction. 
  • The gun can shot a bullet with speed u.
  • Consider a bird is flying in a horizontal line above the man at an altitude y0, as shown in figure.
  • The maximum height which the bullet can attain is

 \(H_{max} =\dfrac{u^2 sin^2\theta}{2g}\)

  • Thus, to hit the bird at altitude \(y_0\), the bird should be at altitude which is equal to or less than Hmax i.e.,

\(y_0 \leq H_{max}\)

 

Case 1

If y0 < Hmax , then bullet can hit the plane at two points A and B, as the bird can attain altitude y0  at two points, as shown in figure.

Case 2 

If y0 = H max , only one point is possible where the bullet can hit the bird, as shown in figure.

 

  • Mathematically the trajectory of a projectile is given as 

           \( y = x \,tan\, \theta - \dfrac{g\,x^2}{2\,u^2 cos^2\,\theta}\)

 Above equation is a quadratic equation in terms of \(x\).

  • Arranging above equation 

\(\left(\dfrac{g}{2\,u^2 cos^2\theta}\right)x^2 - (tan\,\theta)x + y =0\)

Comparing above equation with general quadratic equation 

\(ax^2 + bx +c =0\)

\(a= \dfrac{g}{2\,u^2\,cos^2\theta}\) ,   \( b = –tan\,\theta,\,\,\,\,\, c = y\)

Discriminant

Discriminant is denoted by D and is equal to b2 - 4ac 

         \(D = b^2 – 4ac\)

  • If D > 0 then, two roots of x are possible. 
  • If D = 0 then, one root of x is possible.
  • If D < 0 then, no root of x is possible.

Illustration Questions

A projectile is projected at an angle \(\theta=\) 30º with speed u = 40 m/s. How many values are possible for \(x\) at y= 10 m? \( (g = 10\, m/s^2​)\)

A Three values 

B One value

C Distinct two values 

D No value

×

The equation of trajectory of projectile is given by 

\(y= x \,tan\, \theta - \dfrac{g\,x^2}{2\,u^2 cos^2\theta}\)

Here, \(\theta =30º, \,u= 40\, m/s,\, y = 10\,m\)

\(10 = x \, tan\, 30º - \dfrac{10×x^2}{2(40)^2(cos\,30º )^2}\)

\(10 = x\left(\dfrac{1}{\sqrt3}\right) - \dfrac{x^2}{240}\)

\(\dfrac{x^2}{240}- \dfrac{x}{\sqrt3} + 10 = 0\)        .....(1)

Comparing equation (1) with general quadratic equation

\(ax^2 + bx + c = 0\)

\(a = \dfrac {1}{240}\)\( b = \dfrac {-1}{\sqrt3}\)\(c= 10\)

Discriminant ,  D = b2 – 4ac

D = \(\left(\dfrac{-1}{\sqrt3}\right)^2 - 4 \left(\dfrac{1}{240}\right) (10)\) 

D= \(\dfrac{1}{3} - \dfrac{1}{6}\)

D = \(\dfrac{1}{6}\)

Since, D > 0 

Thus, two distinct values of \(x\) are possible at y= 10 m.

A projectile is projected at an angle \(\theta=\) 30º with speed u = 40 m/s. How many values are possible for \(x\) at y= 10 m? \( (g = 10\, m/s^2​)\)

image
A

Three values 

.

B

One value

C

Distinct two values 

D

No value

Option C is Correct

Sum of Two Roots of Quadratic Equation

  • In projectile motion, there can be two values of x - coordinate for any specific y - coordinate, as shown in figure.
  • Because the projectile attains a specific height two times during a projectile motion, once in upward motion and once in downward motion.

Values at Maximum Height

  • At maximum height, both the \(x-\) coordinates are equal i.e., there is only one value of \(x\) at maximum height.              

  • To understand the sum of two roots of quadratic equation, consider a particle which is projected with an initial speed u at an angle \(\theta\)  with horizontal direction.
  • Consider two points A and B on the trajectory of  the projectile at same height y0, as shown in figure.
  • The horizontal components of velocity at A and B will be equal to \( u \,cos\, \theta\).
  • The vertical components of velocity at A and B will be equal in magnitude \((v_{y_0})\) but opposite in direction as the speed of a  projectile at same height is equal.
  • The time taken by the particle in reaching  from O to A will be equal to the time taken in reaching from B to R as the magnitude of change in velocity in both the situations is same and the magnitude of acceleration is also same and is equal to g.
  • It concludes that the distance from O to P is equal to the distance from Q to R. 

             \( QR = x_1 = OP\)

 

Range

  • Range,  \(R = OQ + QR\)

            Range, \(R = x_2 + x_1\)

  • Thus, sum of two roots of quadratic equation of projectile motion gives the horizontal range.
  • For,  y0 = Hmax  

                   \(R = x_1 + x_2\)                              \([\because\,\,\, x_1 = x_2 = x]\)

                   \(R = 2x\)

  • Mathematically, the trajectory of the particle in a projectile motion which is projected at an angle \(\theta\) with velocity u is given by

                  \( y = x\, tan\,\theta - \dfrac{g\,x^2}{2\,u^2\,cos^2 \theta}\)

          Above equation is a quadratic equation in \(x\).

  On rearranging it, 

     \(x^2 - \dfrac{2\,u^2 x\,cos^2\theta}{g} tan\,\theta\,+ \dfrac{2\,u^2 \,cos^2\theta}{g} y =0\)

     \(x^2 - \dfrac{\,u^2 \,sin\,2\theta}{g} \,x + \dfrac{2\,u^2 \,cos^2\theta}{g} y =0\)    ....(i)

  • Comparing above equation (i) with general quadratic equation 

          \(ax^2 + bx +c =0\)

          \(a=1 \)\(b = \dfrac{- u^2 \, sin \,2 \theta}{g}\)\(c = \dfrac{2 \,u^2 \, cos^2 \theta}{g}y\)

  • The sum of two roots of a quadratic equation can be given by,

Sum of roots  = \(-\dfrac{b}{a}\)

Sum of roots = \(\dfrac{u^2\,sin\,2\theta}{g}\)

Sum of roots = Range

Illustration Questions

A particle is projected in such a manner that it attains some height y0 at horizontal displacement  \(x_1 = \) 50 m . Again, in same trajectory it attains that same height at horizontal displacement \(x_2 = \) 120 m. What will be the range of the projectile motion?

A 150 m

B 170 m

C 120 m

D 50 m

×

Range = Sum of two roots 

Range = \(x_1 + x_2\)

image

Here, \(x_1 = \) 50 m , \(x_2 = \) 120 m

Range = 50 m + 120 m

= 170 m

image

A particle is projected in such a manner that it attains some height y0 at horizontal displacement  \(x_1 = \) 50 m . Again, in same trajectory it attains that same height at horizontal displacement \(x_2 = \) 120 m. What will be the range of the projectile motion?

A

150 m

.

B

170 m

C

120 m

D

50 m

Option B is Correct

Visual Idea of Two Roots

  • Consider a man holding a rifle at point O who wants to hit a target A, as shown in figure. The coordinates of target A are (x0, y0) . The bullet can be fired with speed 'u'.
  • To hit the target, it is necessary that the point A lies on the trajectory of bullet.
  • Two trajectories of bullet are possible, as shown in figure. Point A lies on both trajectories. 
  • Corresponding to these trajectories, there are two angles  \(\theta_1 \)  and  \(\theta_2 \)   i.e., by both angles, bullet can hit the target.
  • Mathematically, if point of projection is considered as origin, horizontal direction as \(x-\)direction and vertical direction as y -direction, then the trajectory of particle will be given as, 

            \(y= x \, tan\, \theta - \dfrac{g\,x^2}{2\,u^2 cos^2 \theta}\)

  • For any specific value of \(x_0,\, y_0 ,\,u\) and \(g\), equation becomes 

\(y_0= x_0 \, tan \,\theta - \dfrac{g\,x_0^2}{2\,u^2 cos^2 \theta}\)

  • Above equation is quadratic in terms of \(\theta\), which gives two values of  \(\theta\).
  • For a specific point, trajectories at two angles are possible.

Illustration Questions

A monkey is sitting on a branch of a tree. A hunter standing at some distance on the ground wants to hit the monkey with an arrow, with speed 'u' as shown in figure. How many values of \(\theta\) are possible, so that monkey would be hit by the arrow?

A One 

B Two

C Three

D Four

×

If point of projection is considered as origin, horizontal direction as x- direction and vertical direction as y-direction, then the trajectory of particle will be given as

 \(y= x \, tan\, \theta - \dfrac{g\,x^2}{2\,u^2 cos^2 \theta}\)

 

 

For any specific value of \(x_0,\, y_0 ,\,u\) and \(g\) equation becomes 

\(y_0= x_0 \, tan\, \theta - \dfrac{g\,x_0^2}{2\,u^2 cos^2 \theta}\)

Above equation is quadratic in terms of \(\theta\) , which gives two values of  \(\theta\)

For a specific point, trajectories at two angles are possible.

A monkey is sitting on a branch of a tree. A hunter standing at some distance on the ground wants to hit the monkey with an arrow, with speed 'u' as shown in figure. How many values of \(\theta\) are possible, so that monkey would be hit by the arrow?

image
A

One 

.

B

Two

C

Three

D

Four

Option B is Correct

Calculation of Angle made by Trajectory

  • Consider a man holding a rifle at point O who wants to hit a target A, as shown in figure . The coordinates of target A are \((x_0, y_0)\). The bullet can be fired with speed 'u'.
  • To hit the target, it is necessary that the point A lies on the trajectory of bullet.
  • Two trajectories of bullet are possible, as shown in figure. Point A lies on both trajectories. 
  • Corresponding to these trajectories, there are two angles \(\theta_1 \) and \(\theta_2 \) i.e., by both angles, bullet can hit the target.
  • Mathematically, if point of projection is considered as origin, horizontal direction as x-direction and vertical direction as y - direction, then the trajectory of particle will be given as, 

            \(y= x \, tan\, \theta - \dfrac{g\,x^2}{2\,u^2 cos^2 \theta}\)

  • For any specific value of \(x_0, \,y_0 ,\,u\) and \(g\), equation becomes 

\(y_0= x_0 \, tan \,\theta - \dfrac{g\,x_0^2}{2\,u^2 cos^2 \theta}\)

  • Above equation is quadratic in terms of \(\theta\), which gives two values of  \(\theta\).
  • For a specific point, trajectories at two angles are possible.

Steps to Obtain Angle made by Trajectory with Horizontal Axis

Step 1: Write the equation of the trajectory of the particle.

Step 2:  Put the values of  \(x,\, y,\,g\,\) and \(u\).

Step 3 : Solve for \(\theta\) and get two values of \(\theta\).

  

Illustration Questions

A gun can fire a bullet with a speed \(u = 4 \,m/s\). A target is at horizontal distance \(x=1 \,m\) and vertical distance \(y= \dfrac{3}{8} m\). Find the values of inclination angle of the gun so that bullet hits the target. \(( g = 10\, m/s^2,\,\,tan^{-1} (2.2) = 65º)\)

A 70º and 30º

B 30º and  60º

C 45º and 65º

D 45º and 85º

×

If the origin is taken as point of projection, x- axis as horizontal direction and y - axis as vertical direction then, equation of trajectory of the particle is given by

\(y = x \, tan\,\theta - \dfrac{g\,x^2}{2\,u^2 cos^2 \theta}\)  

Putting   \(y= \dfrac{3}{8} m\)\(x= 1\,m,\, g = 10 \,m/s^2 , \,u = 4 \,m/s\) 

\(\dfrac{3}{8} = tan\,\theta - \dfrac{10(1)^2}{2× (4)^2 cos^2 \theta}\)

\(\dfrac{3}{8} = tan\,\theta - \dfrac{5}{16} sec^2 \theta\)

\(3 = 8 \,tan\,\theta - \dfrac{5}{2} sec^2 \theta\)

\(3 = 8 \,tan\,\theta - \dfrac{5}{2} (1+tan^2 \theta)\)

\(6 = 16 \,tan\,\theta - 5- 5\,tan^2 \theta\)

\(5 \,tan^2\theta - 16\,tan\,\theta+ 11 = 0\)

\((tan\,\theta - 1) (5 \,tan\,\theta - 11) = 0\)

\((tan \,\theta - 1 ) = 0 \) or \(5 \,tan \,\theta - 11 = 0 \) 

\(tan\,\theta =1\) or  \(tan \theta = \dfrac{11}{5}\)

\(\theta = tan^{-1} (1)\) or  \(\theta = tan^{-1} (2.2)\)

\(\theta=\) 45º or  \(\theta=\)  65º

A gun can fire a bullet with a speed \(u = 4 \,m/s\). A target is at horizontal distance \(x=1 \,m\) and vertical distance \(y= \dfrac{3}{8} m\). Find the values of inclination angle of the gun so that bullet hits the target. \(( g = 10\, m/s^2,\,\,tan^{-1} (2.2) = 65º)\)

A

70º and 30º

.

B

30º and  60º

C

45º and 65º

D

45º and 85º

Option C is Correct

Calculation of Two Roots of Quadratic Equation

Two Roots of Quadratic Equation

  • In projectile motion, there can be two values of x - coordinate for any specific y - coordinate, as shown in figure.
  • Because the projectile attains a specific height two times during a projectile motion, once in upward motion and once in downward motion.

Values at Maximum Height

  • At maximum height, both the x - coordinates are equal i.e., there is only one value of x at maximum height.                     

  • To understand this phenomenon, consider a man who tries to kill the flying bird with a gun at an angle  \(\theta\) with horizontal direction. 
  • The gun can shot a bullet with speed u.
  • Consider a bird is flying in a horizontal line above the man at an altitude y0, as shown in figure.
  • The maximum height which the bullet can attain is  \(H_{max} =\dfrac{u^2 sin^2\theta}{2g}\)
  • Thus, to hit the bird at altitude yo, the bird should be at altitude which is equal to or less than Hmax i.e.,

           \(y_0 \leq H_{max}\)

 

Case 1: 

  • If y0 < Hmax , then bullet can hit the plane at two points A and B, as the bird can attain altitude y0 at two points, as shown in figure.

Case 2 :

 If y0 = H max , only one point is possible where the bullet can hit the bird, as shown in figure.

  • Mathematically the trajectory of a projectile is given as 

           \( y = x \,tan \,\theta - \dfrac{g\,x^2}{2\,u^2 cos^2\theta}\)

Above equation is a quadratic equation in terms of  \(x\)

Arranging above equation 

\(\left(\dfrac{g}{2\,u^2 cos^2\theta}\right)x^2 - (tan\,\theta)x + y =0\)

Comparing above equation with general quadratic equation 

\(ax^2 + bx +c =0\)

\(a= \dfrac{g}{2\,u^2\,cos^2\theta}\) ,   \( b = –tan\,\theta,\,\,\,\,\, c = y\)

Discriminant

Discriminant is denoted by D and is equal to b2 - 4ac 

         \(D = b^2 – 4ac\)

  • If D > 0 then, two roots of x are possible. 
  • If D = 0 then, one root of x is possible .
  • If D < 0 then, no root of x is possible .

Steps to Obtain Values of Horizontal Displacement (x) 

Step 1 : Write the equation of trajectory.

Step 2 : Put the values of \( y, u ,\) \(\theta\) and \(g\).

Step 3 : Solve the quadratic equation for  \(x\) and get two values of  \(x\).

Illustration Questions

A projectile is projected at an angle \(\theta= \) 45º with speed u= 30 m/s. Find the values of horizontal displacement \(x\) at y = 12.5 m. Given : g = 10m/s2 

A 75 m, 15 m 

B 5 m, 100 m 

C 60 m, 20 m 

D 150 m, 75 m 

×

The equation of trajectory of projectile is given by

\(y= x \,tan \,\theta - \dfrac{g\,x^2}{2\,u^2 cos^2\theta}\)

Here, \(\theta = \)\( 45º,\, u= 30\,m/s, \,y = 12.5\, m, \,g = 10 \,m /s^2 \)

  \(12.5 = x \,tan \,45º - \dfrac{(10)(x^2)}{2(30)^2(cos \,45º)^2}\,\)

 

\(12.5 = x \,(1) - \dfrac{x^2}{2×90\left(\dfrac{1}{\sqrt2}\right)^2}\)

\(12.5 = x - \dfrac{x^2}{2× 90× \dfrac{1}{2}}\)

\(\dfrac{x^2}{90} - x + 12.5 =0\)

\((x-75)(x-15) = 0\)

\((x-75)=0 \) or  \(x-15=0\)

\(x=75\,m\)  or  \(x=15\,m\)   

A projectile is projected at an angle \(\theta= \) 45º with speed u= 30 m/s. Find the values of horizontal displacement \(x\) at y = 12.5 m. Given : g = 10m/s2 

image
A

75 m, 15 m 

.

B

5 m, 100 m 

C

60 m, 20 m 

D

150 m, 75 m 

Option A is Correct

Calculation of Maximum Height  for a given Value of Horizontal  Displacement 

  • Consider a skyscraper is on fire. The safe distance from the building is \(x\)

  • A fire extinguishing hose pipe disposes water at speed 'u', as shown in figure.

  • When small  \(\theta_L\)  is taken, water will not reach to the maximum height  for distance \(x\).
  • When large \(\theta_H\) is taken, even then also water will not reach to the maximum height for distance \(x\).
  • Thus, there must be intermediate  \(\theta_I\) ,so that water reaches at maximum height for distance \(x\).
  • To calculate  \(y_{max}\), write the quadratic equation of the trajectory of particle and put discriminant \(\geq\) 0.
  • The equation of trajectory  of the projectile 

               \(y= x \, tan\, \theta - \dfrac{g\,x^2}{2\,u^2cos^2\theta}\)           

Illustration Questions

A building is on fire. A fire extinguishing hose pipe disposes water at a speed u = 10 m/s. The safe distance from building is  \(x\) = 5 m. Calculate the maximum height at which water can be disposed from safe distance. 

A \(4.85 \,m\)

B \(3 \,m\)

C \(3.75 \,m\)

D \(2 \,m\)

×

The equation of trajectory of projectile is given by 

\(y= x\, tan \,\theta - \dfrac{g\,x^2}{2\,u^2cos^2\theta}\)

Here, x= 5 m, g = 10 m/s 2, u = 10 m/s

\(y= 5 (tan\,\theta) - \dfrac{(10)(5)^2}{2(10)^2\,cos^2 \theta}\)

\(y= 5 \,tan\,\theta - \dfrac{5}{4} sec^2 \theta\)

\(y= 5 \,tan\,\theta - \dfrac{5}{4}(1 + tan^2\theta)\)

\(y= 5 \,tan\,\theta - \dfrac{5}{4}- \dfrac{5}{4} tan^2\theta\)

\(\dfrac{5}{4} tan^2\theta - 5 \,tan\,\theta + \left(y+\dfrac{5}{4}\right) =0\)

 

For y to be maximum 

Discriminant  \(\geq\) 0

\(b^2- 4ac \geq 0\)

\((-5)^2 - 4 \left(\dfrac{5}{4}\right) \left(y+\dfrac{5}{4}\right) \geq 0\)

\(25 - 5y - 5 \left(\dfrac{5}{4}\right) \geq 0\)

\(25 - \left(\dfrac{25}{4}\right) \geq 5y\)

\(\dfrac{100-25}{4} \geq 5y\)

\(\dfrac{75}{4×5}\geq y\)

\(\dfrac{15}{4}\geq y\)

\(3.75\,m\geq y_{max}\)

A building is on fire. A fire extinguishing hose pipe disposes water at a speed u = 10 m/s. The safe distance from building is  \(x\) = 5 m. Calculate the maximum height at which water can be disposed from safe distance. 

A

\(4.85 \,m\)

.

B

\(3 \,m\)

C

\(3.75 \,m\)

D

\(2 \,m\)

Option C is Correct

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