Hi Every one. Can any one help me to start solving this problem?i am stocked in the beginning.
The range of a ballistic projectile is given by
r=(v^2)sin2x/g
where v is the initial velocity, x is the launch angle (relative to the horizontal) and g is the
gravitational acceleration. The launch angle cannot be controlled precisely so is treated as a random variable x with f x(x). It should be clear that the maximum range occurs for x=0.25pi
.
#Determine an expression for the PDF of the range (now a random variable,r~f (r))
assuming that x can never exceed the interval 0<x<0.5pi(0 and 0.5pi included)
thanks
Last edited by 22nelson; 03-14-2012 at 08:02 PM.
actually i just found out i need to know the pdf os R which is equal to pdf of x in some point. i hope i did not make any one confuse
thanks for your concern
I think you were saying that you figured it out? I don't know... I'm just confused...
I don't have emotions and sometimes that makes me very sad.
no i did not
so do you think if i start like: P(R<a)
it gives me the probability of R in point (a) which is equal to P(x). then derivative of that to get pdf
am i in right track?
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