Informative line

### 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.

#### 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? $$(g = 10\, m/s^2​)$$

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

#### 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 = $$x_1 + x_2$$

Here, $$x_1 =$$ 50 m , $$x_2 =$$ 120 m

Range = 50 m + 120 m

= 170 m

### 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

# 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  $$x$$.

#### 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

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

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

#### 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

# 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.

#### 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?

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 $$\theta$$.

#### 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