Derivation of area of trajectory of projectile

[Quantum Mind]

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

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

The Attempt at a Solution

I know that tan \Theta = y/x, 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?

 

[TSny]

The equation is derived from y = v_oyt - (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.

 

[TSny]

Homework Statement

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

In the equation you want to derive, \Theta is the initial angle of projection. It is not the angle such that tan \Theta = y/x for a point on the trajectory.

 

[Quantum Mind]

The equation is derived from y = v_oyt - (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

t = 2u sin\theta/g

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

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

I used the equation S = ut - 1/2gt^2 and tried to substitute the value of 't', but I ended up with something else.

 

[Quantum Mind]

OK, I got it.

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

x = ut + at^2/2

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

distance traveled in y-axis is

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

Now I can substitute the value of t = x/u cos \theta and I get the result. Thanks for the help.

 

[TSny]

Nice work!