Chapter 7.7, Problem 50AYU

a.

To determine

Show that for fixed $v_o$ and $\varphi$ the maximum range down the incline is given by $R_{\max }=\frac{v_o^2}{g\left(1-\sin \varphi \right)}$ .

Answer to Problem 50AYU

  $\boxed{R_{\max }=\frac{v_o^2}{g\left(1-\sin \varphi \right)}}$

Explanation of Solution

Given information:

The range $R$ of a projectile propelled downward from the top of an inclined plane at angle $\theta$ to the inclined plane is given by,

  $R\left(\theta \right)=\frac{2v_o^2\sin \theta \cos \left(\theta -\varphi \right)}{g{\cos }^2\varphi }$ Where $v_o$ the initial velocity of the projectile is, $\varphi$ is the angle the plane makes with respect to the horizontal, and $g$ is acceleration due to gravity.

Show that for fixed $v_o$ and $\varphi$ the maximum range down the incline is given by $R_{\max }=\frac{v_o^2}{g\left(1-\sin \varphi \right)}$ .

Calculation:

Consider the expression,

  $R\left(\theta \right)=\frac{2v_o^2\sin \theta \cos \left(\theta -\varphi \right)}{g{\cos }^2\varphi }$

Apply product-to-sum formula,

  $\sin \alpha \cdot \cos \beta =\frac{1}{2}\left[\sin \left(\alpha +\beta \right)+\sin \left(\alpha -\beta \right)\right]$

  $\begin{array}{l}R\left(\theta \right)=\frac{2v_o^2\sin \theta \cos \left(\theta -\varphi \right)}{g{\cos }^2\varphi }=\frac{2v_o^2\frac{1}{2}\left[\sin \left(\theta +\theta -\varphi \right)+\sin \left(\theta -\left(\theta -\varphi \right)\right)\right]}{g{\cos }^2\varphi }\\ =\frac{v_o^2\left[\sin \left(2\theta -\varphi \right)+\sin \varphi \right]}{g{\cos }^2\varphi }\\ now\\ \sin \left(2\theta -\varphi \right)=1\\ =\frac{v_o^2\left(1+\sin \varphi \right)}{g{\cos }^2\varphi }\\ =\frac{v_o^2\left(1+\sin \varphi \right)}{g\left(1-{\sin }^2\varphi \right)}=\frac{v_o^2\left(1+\sin \varphi \right)}{g\left(1+\sin \varphi \right)\left(1-\sin \varphi \right)}\\ R_{\max }=\frac{v_o^2}{g\left(1-\sin \varphi \right)}\end{array}$

Hence, $\boxed{R_{\max }=\frac{v_o^2}{g\left(1-\sin \varphi \right)}}$ .

b.

To determine

Determine the maximum range if the projectile has an initial velocity of $50m/\sec$ ,the angle of the plane is $\varphi =35°$ and $g=9.8m/{\sec }^2$ .

Answer to Problem 50AYU

  $R_{\max }=\boxed{598meters}$

Explanation of Solution

Given information:

The range $R$ of a projectile propelled downward from the top of an inclined plane at angle $\theta$ to the inclined plane is given by,

  $R\left(\theta \right)=\frac{2v_o^2\sin \theta \cos \left(\theta -\varphi \right)}{g{\cos }^2\varphi }$ Where $v_o$ the initial velocity of the projectile is, $\varphi$ is the angle the plane makes with respect to the horizontal, and $g$ is acceleration due to gravity.

Determine the maximum range if the projectile has an initial velocity of $50m/\sec$ ,the angle of the plane is $\varphi =35°$ and $g=9.8m/{\sec }^2$ .

Show that for fixed $v_o$ and $\varphi$ the maximum range down the incline is given by $R_{\max }=\frac{v_o^2}{g\left(1-\sin \varphi \right)}$ .

Calculation:

Consider the expression,

  $R_{\max }=\frac{v_o^2}{g\left(1-\sin \varphi \right)}$

Substitute the given values in above expression, we get

  $\begin{array}{l}{R}_{\max }=\frac{v_o^2}{g\left(1-\sin \varphi \right)}=\frac{{50}^2}{9.8\left(1-\sin 35°\right)}\\ =598meters\end{array}$

Hence, $R_{\max }=\boxed{598meters}$ .