Solving initial velocity problems
Author
I'm desperately looking for a solution to this problem. No, it doesn't pertain to game physics, well, I'm certain it does but any help for a budding game programmer will be greatly appreciated.
For the life of me, I cannot figure out how to solve a problem for initial velocity involving non-horizontal projectile motion. How can you find the initial velocity of a vector with the just the angle of projection and it's target coordinates?
Again, any help would be greatly appreciated :)
Is this homework? ...
Known:
d (the displacement), θ (the firing angle), g (global gravity)
Want:
v (initial velocity)
Combining θ and v using trig. we have a relationship between the two initial velocities in the x and y axes vx and vy:
vx = v cos θ
vy = v sin θ
Now the only component of the velocity that changes due to vertical acceleration due to gravity is the y component. Thus the x component (vx) is constant over the time it takes to travel d:
t vx = d
therefore vx = d/t
and then v cos θ = d/t
As for the y component, when the projectile lands it will have the same speed downwards as it did upwards to begin with, which means that:
vy' = vy - 1/2 g t2 = -vy
therefore 2vy = 1/2 g t2
and thus v sin θ = 1/4 g t2
The two things we don't know in this mess is v and t. But we have two equations. It's just algebra from here.
[Edited by - Doc on February 10, 2005 1:44:11 AM]
| /|| v / || / || / ||/) θ |+---------------------------------x-------| d |Known:
d (the displacement), θ (the firing angle), g (global gravity)
Want:
v (initial velocity)
Combining θ and v using trig. we have a relationship between the two initial velocities in the x and y axes vx and vy:
vx = v cos θ
vy = v sin θ
Now the only component of the velocity that changes due to vertical acceleration due to gravity is the y component. Thus the x component (vx) is constant over the time it takes to travel d:
t vx = d
therefore vx = d/t
and then v cos θ = d/t
As for the y component, when the projectile lands it will have the same speed downwards as it did upwards to begin with, which means that:
vy' = vy - 1/2 g t2 = -vy
therefore 2vy = 1/2 g t2
and thus v sin θ = 1/4 g t2
The two things we don't know in this mess is v and t. But we have two equations. It's just algebra from here.
[Edited by - Doc on February 10, 2005 1:44:11 AM]
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