I think it is atan of the derivative of the curve, is that right?
The derivative is the slope of the tangent line. The atan is thus the angle of the tangent line with the x axis.
Thanks for the quick reply. Yeah, that is what I thought. However, I have a followup question.
For the vertical position of a projectile the formula is:
y = initialheight + VyT - .5(9.8)T^2
The derivative of that is
d/dx = Vy - 9.8T
I did a sample problem, Vi=4, angle = 30º, initial height = 1.2
Total Time: .739 seconds
Height of Max time: .202 seconds
Distance: 2.56 m
if I take atan( Vy - 9.8T) when t = 0 the angle comes up as 63º, not 30º which it should be.
I started to work out the problem, but you didn't state it quite clearly....
Do you mean V0x is the initial Vi in the direction of vector i ?
dy/dx is the arctan of the angle the curve makes with the horizontal.
dy/dt has no geometric meaning (in an xy-coordinate system).
If Vi= 4 is the initial speed and the projectile is fired at a 30 degree angle to the horizontal, then the initial vertical speed is Vy= 4 cos(30)= 4(1/2)= 2 and the initial horizontal speed is
Vx= 4 sin(30)= 4([sqrt](3)/2)= 2 [sqrt](3).
The vertical speed would be given by dy/dt= 2- 9.8t and the horizontal speed (assuming gravity is the only force) by
dx/dt= 2[sqrt]3. At t= 0, dy/dt= 2 and dx/dt= 2[sqrt](3).
By the chain rule, dy/dx= (dy/dt)/(dx/dt)= 2/2[sqrt](3)= 1/[sqrt](3)= [sqrt](3)/3, the tangent of 30 degrees.
Thanks, Hallsofivy. You created a good thorough reply. However, I think you accidentally swapped cosine and sine in Vx and Vy.
Anyway, for the benifit of searches, here is the formula for the angle:
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