Derivation of area of trajectory of projectile

In summary, the equation relates y to x (horizontal range) and t (time of flight) in terms of projectile motion.
  • #1
Quantum Mind
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Homework Statement



This is a problem about an aircraft flying above a cannon which is capable of firing in any direction. The plane's height, velocity and the speed of the projectile fired by the cannon are given (I am not trying to get help in solving the problem, so the numbers are not given). I am supposed to find the time duration when the plane is at risk of get hit. I understood the problem, but my query is about the derivaiton of the formula for getting the area of the projectile's trajectory

Homework Equations



[tex]y = x tan \Theta - gx^2 / 2 u^2 cos^2\Theta[/tex]

The Attempt at a Solution



I know that [tex]tan \Theta = y/x[/tex], but why are we reducing the second component from this area? Does this have something do with the shape of the path of the projectile being a parabola?
 
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  • #2
The equation is derived from y = voyt - (1/2)gt2.

Since you are trying to get an equation relating y to x rather than y to t, you need to find an expression for t in terms of x and substitute it into this equation. So, think about how x and t are related in projectile motion.
 
Last edited:
  • #3
Quantum Mind said:

Homework Statement



I know that [tex]tan \Theta = y/x[/tex],...

In the equation you want to derive, [itex]\Theta[/itex] is the initial angle of projection. It is not the angle such that [itex]tan \Theta = y/x[/itex] for a point on the trajectory.
 
  • #4
TSny said:
The equation is derived from y = voyt - (1/2)gt2.

Since you are trying to get an equation relating y to x rather than y to t, you need to find an expression for t in terms of x and substitute it into this equation. So, think about how x and t are related in projectile motion.

time of flight

[tex]t = 2u sin\theta/g[/tex]

x (horizontal range) = [tex]u^2 sin2\theta / g[/tex]

How is this related to [tex] gx^2 / 2u^2cos^2\Theta ?[/tex]

I used the equation [tex]S = ut - 1/2gt^2[/tex] and tried to substitute the value of 't', but I ended up with something else.
 
  • #5
OK, I got it.

For a projectile motion, the distance traveled in x-axis is

[tex]x = ut + at^2/2[/tex]

But since 'a' is zero along the x axis, [tex]x = ucos\theta.t[/tex]

distance traveled in y-axis is

[tex]y = u sin\theta.t - 1/2 gt^2[/tex].

Now I can substitute the value of [tex]t = x/u cos \theta[/tex] and I get the result. Thanks for the help.
 
  • #6
Nice work! :cool:
 

Related to Derivation of area of trajectory of projectile

1. What is the derivation of the area of trajectory of projectile?

The derivation of the area of trajectory of projectile is a mathematical process that involves using the equations of motion to determine the area covered by a projectile in flight. This calculation is important in understanding the path and distance traveled by a projectile.

2. How is the area of trajectory of projectile calculated?

The area of trajectory of projectile is calculated using the formula A = (Vx * Vy * sin(2θ))/g, where Vx is the horizontal velocity, Vy is the vertical velocity, θ is the angle of launch, and g is the acceleration due to gravity. This formula takes into account the time of flight, the initial velocity, and the angle of launch to determine the area covered by the projectile.

3. Why is the derivation of the area of trajectory of projectile important?

The derivation of the area of trajectory of projectile is important in understanding the motion of projectiles and their paths. This information is useful in fields such as physics, engineering, and ballistics, where the trajectory of a projectile is crucial in predicting its behavior and impact.

4. Are there any assumptions made in the derivation of the area of trajectory of projectile?

Yes, there are a few assumptions made in the derivation of the area of trajectory of projectile. These include the absence of air resistance, a constant acceleration due to gravity, and a flat and uniform surface for the projectile to travel on. Real-life situations may vary from these assumptions, but they provide a good approximation for calculations.

5. Can the derivation of the area of trajectory of projectile be applied to all types of projectiles?

Yes, the derivation of the area of trajectory of projectile can be applied to all types of projectiles as long as the assumptions mentioned above hold true. This includes projectiles launched at different angles, with different initial velocities, and in different environments. However, it is important to note that the calculation may not be entirely accurate in some real-life scenarios due to variations in the assumptions made.

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