Question

41. Find the distribution of R-A sin θ, where A is a fixed constant and θ is uniformly distributed on (-π/2, π/2). Such a random variable R arises in the theory of ballistics. If a projectile is fired from the origin at an angle α from the earth with a speed v, then the point R at which it returns to the earth can be expressed as R--(W/g) sin 2α, where g is the gravitational constant, equal to 980 centimeters per second squared.

Answer 1

SOLUTION:

From given data,

(41).Given $\theta$ is uniformly distributed on (-$\pi$/2 , $\pi$/2)

$f_{\theta }$ ($\theta$) = 1/$\pi$ (-$\pi$/2 < $\theta$< $\pi$/2)  

$f_{\theta }$ ($\theta$) = 0 otherwise

R = A Sin $\theta$ ,

Where A is constant

Consider the C.D.F of R

(r) = P(R < r)

= P(A Sin $\theta$  < r)

= P($\theta$< Sin^-1 (r /A) )

=

Sin-(r/A) (1/T)do

(1/T) * θ|sin/2(r/A)

= 1/$\pi$ * (sin^-1 (r/A) + ($\pi$/2))

fR
(r) = d/dr   (r)

= 1/$\pi$ * d/dr (sin^-1 (r/A)) + 0

= 1/$\pi$ * (1/sqrt(1-(r/A)²))*(1/A))

fR
(r) = 1/$\pi$ * (1/sqrt(A² - r²)) - A < r < A

As we know that

V = v + at

Where,

V = final velocity (0)

v = initial velocity

0 = v sin $\alpha$ - gt

t = v sin $\alpha$ / g

As Ascend = Descend

Total

T₀ = 2v sin $\alpha$ / g

Range = Horizontal velocity * time of flight

Range = v *T₀

Range = ( v cos $\alpha$ * 2 v sin $\alpha$ ) / g

Range = ( 2 v² sin $\alpha$ cos $\alpha$) / g

Range = R = (v² sin 2$\alpha$) / g

Please thumbs-up / vote up this answer if it was helpful. In case of any problem, please comment below. I will surely help. Down-votes are permanent and not notified to us, so we can't help in that case.