projectile launch velocities, given a position

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May 5th 2008, 03:32 AM
smitzer

projectile launch velocities, given a position

struggling with this problem:
using very simple projectile rules (no air resistance ect) and only gravity acting on the object, is it possible to:
with a known: gravity, launch magnatude and coodinates which the partical passes through, to work out the launch angle (or x,y components?)

using the position equations:
X=Vx*t
Y=Vy*t+(g/2)*t^2

how do i find Vx and Vy?
itld be easy if i knew time, but i dont, however i do know magnatude.
sqrt(Vx^2+Vy^2)=M

anyone got any ideas?
May 5th 2008, 03:35 AM
topsquark

Quote:

Originally Posted by smitzer

struggling with this problem:
using very simple projectile rules (no air resistance ect) and only gravity acting on the object, is it possible to:
with a known: gravity, launch magnatude and coodinates which the partical passes through, to work out the launch angle (or x,y components?)

using the position equations:
X=Vx*t
Y=Vy*t+(g/2)*t^2

how do i find Vx and Vy?
itld be easy if i knew time, but i dont, however i do know magnatude.
sqrt(Vx^2+Vy^2)=M

anyone got any ideas?

$Vx = M~cos(\theta)$
and
$Vy = M~sin(\theta)$
where $\theta$ is the launch angle.

You'll have two equations in two unknowns which you can solve. My recommendation is to solve the x equation for cos(theta) and the y equation for sin(theta), then use
$sin^2(\theta) + cos^2(\theta) = 1$
to get an equation for t. (It's a biquadratic which you can solve for t^2 using the quadratic formula. Then solve for t from there.)

FYI: You will get two angles.

-Dan
May 5th 2008, 04:45 AM
smitzer

thanks dan,(thats my name also (Cool))
so if:
$cos(\theta)=x/Mt$
and
$sin(\theta)=(y-(g/2)*t^2)/Mt$
i should be able to find t through:
$((y-(g/2)*t^2)/Mt)^2+(x/Mt)^2=1$
is that right?
May 5th 2008, 06:16 AM
topsquark

Quote:

Originally Posted by smitzer

thanks dan,(thats my name also (Cool))
so if:
$cos(\theta)=x/Mt$
and
$sin(\theta)=(y-(g/2)*t^2)/Mt$
i should be able to find t through:
$((y-(g/2)*t^2)/Mt)^2+(x/Mt)^2=1$
is that right?

Looks good to me.

These are nasty, aren't they? I always used to scare my students by doing one of these as an example. (Evilgrin) Just to show them that I wasn't being mean by giving them the "hard" problems for homework. Really. That was the only reason. (Itwasntme)

-Dan
May 5th 2008, 07:33 AM
smitzer

i got
$(g/2)*t^4-(M-g)*t^2+(x^2+y^2)=0$
but when i try some example numbers:
g=0.3
m=0.5
x,y=10; ect
my discriminant is always negative.

i may have done that step wrong, i havent warmed up my maths brain in ages. (Headbang)

again, thanks for any help :)
May 5th 2008, 07:50 AM
topsquark

Quote:

Originally Posted by smitzer

i got
$(g/2)*t^4-(M-g)*t^2+(x^2+y^2)=0$
but when i try some example numbers:
g=0.3
m=0.5
x,y=10; ect
my discriminant is always negative.

Two comments. First, g = 9.8 m/s^2 so you might as well use that value when cranking out the numbers.

Also, I owe you an apology.

Quote:

Originally Posted by smitzer

$sin(\theta)=(y-(g/2)*t^2)/Mt$

This equation is not quite right. It should be
$sin(\theta)=(y+(g/2)*t^2)/Mt$

Sorry I didn't catch that earlier.

-Dan
May 6th 2008, 06:47 AM
smitzer

solved this in the end :)
i used the sin and cos to make tan,
giving me a quadratic equation for it, with 4 answers

thanks for the help, you certainly steered me in the right direction
(Rofl)