Q 129. - Precalculus Enhanced with Graphing Utilities

Question

Projectile Motion - An object is propelled upward at an angle $\theta$ between $45°$and $90°$ to the horizontal with an initial

velocity of ${\nu }_o$ feet per second from the base of an inclined plane that makes an angle of 45° with the horizontal. See the

illustration. If air resistance is ignored, the distance R that it travels up the inclined plane as a function of $\theta$ is given by

$R\left(\theta \right)=\frac{{{\nu }_o}^2\sqrt{2}}{32}\left[\sin \left(2\theta \right)-\cos \left(2\theta \right)-1\right]$

(a) Find the distance R that the object travels along the inclined plane if the initial velocity is 32 feet per second and $\theta =60°$

(b) Graph R = R$\left(\theta \right)$ if the initial velocity is 32 feet per second.

(c) What value of $\theta$ makes R largest?

Step-by-Step Solution

Answer

(a) Distance R is $16\sqrt{2}\left(\sqrt{3}-1\right)$.

(c) $\theta =60°$ makes R largest.

Step 1. Given information.

Consider the given angle$\theta =60°$ and velocity  32 feet per second.

Step 2. Finding Distance.

Substitute the values $\theta =60°$and initial velocity 32 feet per second in  given formula .

$$\begin{gathered} R\left(\theta \right)=\frac{{\nu }_{o}2\sqrt{2}}{32}\left[\sin \left(2\theta \right)-\cos \left(2\theta \right)-1\right] \\ R\left(60°\right)=\frac{{\left(32\right)}^2\sqrt{2}}{32}\left[\sin 120°-\cos 120°-1\right] \\ =16\sqrt{2}\left(\sqrt{3}-1\right) \end{gathered}$$