Q 5.41 - A First Course in Probability

Question

Find the distribution of$R=A\sin \left(\theta \right)$, where $A$is a fixed constant and $\theta$is uniformly distributed on$\left(\raisebox{1ex}{$-\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)$. Such a random variable $R$arises in the theory of ballistics. If a projectile is fired from the origin at an angle $\alpha$from the earth with a speed$\nu$, then the point $R$at which it returns to the earth can be expressed as$R=\left(\raisebox{1ex}{$v^2$}\!\left/ \!\raisebox{-1ex}{$g$}\right.\right)\sin \left(2\alpha \right)$, where $g$is the gravitational constant, equal to $980$centimeters per second squared. 

Step-by-Step Solution

Answer

Therefore, it returns to the earth can be expressed as $R=\left(\raisebox{1ex}{$v^2$}\!\left/ \!\raisebox{-1ex}{$g$}\right.\right)\sin \left(2\alpha \right)$

Step 1 Given information:

$\theta$is uniformly distributed on$\left(\raisebox{1ex}{$-\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)$

Step 2 Explanation:

$f_{\theta }\left(\theta \right)=\frac{1}{\mathrm{\pi }}\left(-\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\le \theta \le \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)$

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

$R=A\sin \left(\theta \right)$

Where $A$is constant

consider the C.D.F of $R$

$$\begin{gathered} F_R\left(r\right)=P(R\le r) \\ =P(A\sin \left(\theta \right)\le r) \\ =P(\theta \le {\sin }^{-1}\left(\raisebox{1ex}{$r$}\!\left/ \!\raisebox{-1ex}{$A$}\right.\right)) \\ ={\int }_{\raisebox{1ex}{$-\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{{\sin }^{-1}\left(\raisebox{1ex}{$r$}\!\left/ \!\raisebox{-1ex}{$A$}\right.\right)}\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\mathrm{\pi }$}\right.\right)d\theta \\ =\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\mathrm{\pi }$}\right.\right)\times {\overline{)\theta }}_{\raisebox{1ex}{$-\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{{\sin }^{-1}\left(\raisebox{1ex}{$r$}\!\left/ \!\raisebox{-1ex}{$A$}\right.\right)} \\ =\frac{1}{\mathrm{\pi }}\times \left({\sin }^{-1}\left(\raisebox{1ex}{$r$}\!\left/ \!\raisebox{-1ex}{$A$}\right.\right)+\left(\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)\right) \\ {f}_R\left(r\right)=\frac{d}{dr}{F}_R\left(r\right) \\ =\frac{1}{\mathrm{\pi }}\times \frac{d}{dr}\left({\sin }^{-1}\left(\raisebox{1ex}{$r$}\!\left/ \!\raisebox{-1ex}{$A$}\right.\right)\right)+0 \\ =\frac{1}{\mathrm{\pi }}\times \left(\frac{1}{\sqrt{1-{\left({\displaystyle \raisebox{1ex}{$r$}\!\left/ \!\raisebox{-1ex}{$A$}\right.}\right)}^2}}\times \frac{1}{A}\right) \\ {f}_R\left(r\right)=\frac{1}{\mathrm{\pi }}\times \left(\frac{1}{\sqrt{A^2-r^2}}\right)-A\le r\le A \end{gathered}$$

Step 3 Explanation:

As we know that

$V=v+at$

Where,

$V=$Final velocity$\left(0\right)$

$v=$initial velocity

$$\begin{gathered} 0=v\sin \left(\alpha \right)-gt \\ t=\frac{v\sin \left(\alpha \right)}{g} \end{gathered}$$

As Ascend $-$ Descend

Total

$T_{{}_0}=\frac{2v\sin \left(\alpha \right)}{g}$

Range$=$Horizontal velocity$\times$time of flight

Range$=v\times T_0$

$$\begin{gathered} =\frac{\left(v\cos \left(\alpha \times 2v\sin \left(\alpha \right)\right)\right)}{g} \\ =\frac{\left(2v^2\sin \left(\alpha \right)\cos \left(\alpha \right)\right)}{g} \\ R=\frac{\left(v^2\sin \left(2\alpha \right)\right)}{g} \end{gathered}$$