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A trajectory or flight path in physics is defined as the path that an object in motion having some mass follows through space as a function of time.Â Hamiltonian mechanics via canonical coordinates is used to define the trajectory of an object in classical mechanics. Hence both position and momentum simultaneously define a complete trajectory. For example, the motion of a ball or a big rock thrown upwards or the motion of a satellite or a bullet fired from a gun .

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The equation of trajectory derivation is as follows ---Â

Â y= x tanÏ´ - (gx2/2 v2 cos2Ï´)Â Â

where, â€˜xâ€™ is the horizontal component of the trajectory.Â

â€˜yâ€™ is the vertical component of the trajectory.

â€˜gâ€™ is the gravitational force of attraction.

â€˜vâ€™ is the initial velocity of the object.

â€˜Ï´â€™ is the angle of elevation of the trajectory.

Time of Flight, T = (2 vo sinÏ´/g)

Maximum Height Reached, H =( v02 sin2Ï´/ 2g )

Horizontal Range , R = (vo2 sin2Ï´/ g)

Where , â€˜voâ€™ is the initial velocityÂ

â€˜sinÏ´â€™ is the vertical component of y-axis

â€˜cosÏ´â€™ is the horizontal component of x-axis.

Thus the trajectory equation along with some important formulae has been derived.

Objects will follow a vertically symmetric path when projected from and land on the same horizontal surface. The flight time depends on the initial velocity of the projectile and the angle of projection. The maximum height of the projectile is reached when the velocity of the object is zero.The range of the projectile depends on the initial velocity of the object.

This article explains the trajectory formula and the derivation of the equation of trajectory. Some FAQs have been added for a better understanding of the topic for the students.Â

FAQ (Frequently Asked Questions)

1. Prove that a gun will shoot three times as high when its angle of elevation is as when it is but covers the same horizontal range.

LetÂ â€˜Ï´â€™ be the angle of elevation.Â

When Ï´=60^{0} , then maximum height ,

H_{1} = (u^{2} sin^{2} 60^{0})/2g = 3u^{2}/8g , where â€˜uâ€™ is the initial velocity of bullet , â€˜gâ€™ is the gravitational force of attraction .Â

Horizontal Range ,

R_{1} = (u^{2} sin(2 X 60^{o}))/g =âˆš3u^{2}/ 2gÂ

When Ï´=30o , then maximum height ,

H_{2}= (u^{2} sin^{2} 300)/2g = u^{2}/8g

Horizontal Range ,Â

R_{2}=Â (u^{2} sin(2 X 30o))/g =âˆš3u^{2}/ 2gÂ

Hence , H_{1} = 3H_{2} Â and R_{1}= R_{2} .

Therefore aÂ gun will shoot three times as high when its angle of elevation is as when it is but covers the same horizontal range.

2. What are the applications and give two examples of the projectile trajectory equation?

A projectile is any object that is cast, fired, flung, heaved, hurled,pitched, tossed, or thrown.Â

**Â Some applications of the projectile trajectory equation in real life are as follows :-Â **

Playing sports like basketball and football as a curve is generated when the ball is thrown or kicked.

Space Explorations require the help of the trajectory formula.

During situations of war and flood operations also the trajectory formula is essential.

**Examples of the projectile trajectory -Â **

A bomb is dropped from a plane during war.

An athlete throwing a javelin .

A ball is kicked or thrownÂ as in football, basketball and cricket respectively.

3. What is the change in momentum between the initial and final points of the trajectory path, if the range is maximum?

If we know that the range is maximum. The angle of projection , Ï´=45^{o} .

Now, since the initial velocity horizontally remains the same throughout the horizontal part, there is no change in momentum in the horizontal.Â Â Â

Magnitude of vertical velocity = u sin 45^{o} ,where â€˜uâ€™ is the initial velocity of the object.Â The vertical velocity will be equal in magnitude but opposite in direction at the required points.

Hence , the change in momentum = 2mu * sin 45^{o} = âˆš2muÂ , where â€˜mâ€™= mass of object. So the change in momentum will be the product of the square of two and mass and initial velocity.Â