Those are two parametric equations. We called them that becuse they are both expressed in terms of t.

To graph them in that form you will need a parametric plotter.

]]>velocity is 10 m/s

at 10 m/s it travels approximately 489 m before touching the ground, and this is quite a large distance (non practical )

I think there is no drag , that is why the length of projectile is so long.

Could you help me inserting drag in this formula

x = vt cos ?

y = vt sin ? - ½ g t²

here you can find some formula for drag, http://en.wikipedia.org/wiki/Drag_%28physics%29

and let me tell you i am little dumb in mathematics

Thanks

EDIT : I want to draw the path of projectile.

]]>with a initial velocity of 10

10 what?

angle it goes to 489 m

Maximum height? Distance? Arc length? 489 m of what?

]]>sorry for reviving this thread again smile

here in this formula

x = vt cos ?

y = vt sin ? - ½ g t²v (velocity) is the initial velocity or a constant velocity ? will velocity changes with time ? or it remains constant ?

In my last post i ask this.

here is the projectile i get using this formula http://imgur.com/axCKwsP

Don't you guys think for 60* angle this is Quite a large projectile ?

with a initial velocity of 10 and 60* angle it goes to 489 m, let me tell you i am decreasing the initial velocity and increasing time by 0.1 at each step.

I dont know i am doing it in right way or not

can any one suggest me any thing ?

here in this formula

x = vt cos θ

y = vt sin θ - ½ g t²

v (velocity) is the initial velocity or a constant velocity ? will velocity changes with time ? or it remains constant ?

]]>Your are welcome and welcome to the forum!

]]>Hi;

To get rid of the 1 / 2 you multiply by 2, because 2 * 1 / 2 = 1

OK I understand now. Thank you very much!

]]>To get rid of the 1 / 2 you multiply by 2, because 2 * 1 / 2 = 1

]]>Hi;

I don't get why 499.6 was divided into 1/2

Both sides were multiplied by -2

This isn't clear to me a bit; why we multiplied both sides by 2? Is it because we had two T variables there has a coefficient of 2? Sorry, my algebra is really that rusted.

]]>I don't get why 499.6 was divided into 1/2

Both sides were multiplied by -2 and some rounding was done

-499.6 * -2 ≈ 999

Then both sides were divided by 9.81 and some rounding done again to get.

100 = t^2

This gives a rough and ready answer for the post.

]]>300 = vt (0.5)

20 = vt (0.866) - ½ g t²the first equation gives us: vt = 300 / 0.5 = 600

put that into the second equation: 20 = 600 x 0.866 - ½ g t²

rearrange: 20 - 600 x 0.866 = - ½ g t²

simplify: 499.6 = ½ g t²

simplify: 999 = g t²

simplify: 999 = t²

solve: t = ± 10 seconds

Hi, sorry for reviving an ANCIENT thread... I have problems understanding the processes done in this equation.

I don't get why 499.6 was divided into 1/2 and g value is omitted? And lastly, how did sqrt 999 became 9.999 or 10?

Once again I am sorry, but once I get all the answers It'll help me tons.

Thanks in advance

]]>I have the following values in my 3d world:

- my current x & y angle

- the distance in meters to the target

- the height offset in meters

- the velocity of the projectile m/s

I need to calculate the angle y offset I need to add to my current angle y so that the missile will hit the target.

I have tried to make a formula but im not good at maths, Im getting confused with the above formulas too because they have t (time), but since my projectile hasn't been fired i won't have time.

Thanks for any help!

]]>x = vt cos θ

y = vt sin θ - ½ g t²

From the first one, get t on its own and sub it into the second one. You don't need to take into account what the time is for your problem. The math would get a bit ugly on this, but after simplifying you should get;

y = x tanθ - gx²(1+ tan² θ)

2v²

So if you need to hit a target x metres away and y metres high, you can work out the angle needed for the velocity you launch it at. Or if you launch it at an angle theta, velocity v and you want to know how far away the projectile is when it is one meter high...

]]>