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...
]]>