an object is propelled upward at an angle θ, 45° < θ < 90°, to the horizontal with an initial velocity of v0…

an object is propelled upward at an angle θ, 45° < θ < 90°, to the horizontal with an initial velocity of v0 feet per second from the base of a 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 is given by r(θ) = v0²√2 / 16 cos θ(sin θ - cos θ). use the given information to answer parts (a) to (d). = v0²√2 / 32 (sin 2θ - (cos 2θ + 1)) use double - angle formulas. = v0²√2 / 32 sin(2θ) - cos(2θ) - 1 use the distributive property. (b) find the angle θ that maximizes r by solving the equation sin(2θ) + cos(2θ) = 0. solve this equation for θ. θ = 67.5° (type an integer or a decimal.) (c) what is the maximum distance r if v0 = 48 feet per second? feet (round to two decimal places as needed.)

an object is propelled upward at an angle θ, 45° < θ < 90°, to the horizontal with an initial velocity of v0 feet per second from the base of a 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 is given by r(θ) = v0²√2 / 16 cos θ(sin θ - cos θ). use the given information to answer parts (a) to (d). = v0²√2 / 32 (sin 2θ - (cos 2θ + 1)) use double - angle formulas. = v0²√2 / 32 sin(2θ) - cos(2θ) - 1 use the distributive property. (b) find the angle θ that maximizes r by solving the equation sin(2θ) + cos(2θ) = 0. solve this equation for θ. θ = 67.5° (type an integer or a decimal.) (c) what is the maximum distance r if v0 = 48 feet per second? feet (round to two decimal places as needed.)

Answer

Explanation:

Step1: Identify the formula for R

We are given $R(\theta)=\frac{v_{0}^{2}\sqrt{2}}{16}\cos\theta(\sin\theta - \cos\theta)$. We know from part (b) that the maximum - value of $R$ occurs at $\theta = 67.5^{\circ}$, and $v_{0}=48$ feet per second.

Step2: Substitute the values into the formula

Substitute $v_{0} = 48$ and $\theta=67.5^{\circ}$ into the formula for $R$. First, calculate the trigonometric values: $\cos(67.5^{\circ})=\frac{\sqrt{2-\sqrt{2}}}{2}$ and $\sin(67.5^{\circ})=\frac{\sqrt{2 + \sqrt{2}}}{2}$. [ \begin{align*} R&=\frac{v_{0}^{2}\sqrt{2}}{16}\cos\theta(\sin\theta - \cos\theta)\ &=\frac{48^{2}\sqrt{2}}{16}\cos(67.5^{\circ})(\sin(67.5^{\circ})-\cos(67.5^{\circ}))\ \end{align*} ] [ \begin{align*} \sin(67.5^{\circ})-\cos(67.5^{\circ})&=\frac{\sqrt{2+\sqrt{2}}}{2}-\frac{\sqrt{2 - \sqrt{2}}}{2}\ &=\frac{\sqrt{2+\sqrt{2}}-\sqrt{2 - \sqrt{2}}}{2} \end{align*} ] [ \begin{align*} \frac{48^{2}\sqrt{2}}{16}&=\frac{2304\sqrt{2}}{16}=144\sqrt{2} \end{align*} ] [ \begin{align*} R&=144\sqrt{2}\times\frac{\sqrt{2-\sqrt{2}}}{2}\times\frac{\sqrt{2+\sqrt{2}}-\sqrt{2 - \sqrt{2}}}{2}\ \end{align*} ] Using the identity $(a + b)(a - b)=a^{2}-b^{2}$, we know that $\sqrt{2+\sqrt{2}}\times\sqrt{2-\sqrt{2}}=\sqrt{4 - 2}=\sqrt{2}$. [ \begin{align*} R&=144\sqrt{2}\times\frac{\sqrt{2-\sqrt{2}}}{2}\times\frac{\sqrt{2+\sqrt{2}}-\sqrt{2 - \sqrt{2}}}{2}\ &=36(\sqrt{2+\sqrt{2}}\sqrt{2-\sqrt{2}}-\left(2 - \sqrt{2}\right))\ &=36(\sqrt{2}-2 + \sqrt{2})\ &=36(2\sqrt{2}-2)\ &=72(\sqrt{2}-1)\ &\approx72(1.414 - 1)\ &=72\times0.414\ & = 29.81 \end{align*} ]

Answer:

$29.81$