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). (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). (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: Recall the formula for $R(\theta)$

We know that $R(\theta)=\frac{v_{0}^{2}\sqrt{2}}{16}\cos\theta(\sin\theta - \cos\theta)$. First, we simplify it to $R(\theta)=\frac{v_{0}^{2}\sqrt{2}}{32}(\sin2\theta-(\cos2\theta + 1))$ using double - angle formulas ($\sin2\alpha = 2\sin\alpha\cos\alpha$ and $\cos2\alpha=2\cos^{2}\alpha - 1$). The angle $\theta$ that maximizes $R$ is $\theta = 67.5^{\circ}$.

Step2: Substitute values into the formula

We are given $v_{0}=48$ feet per second and $\theta = 67.5^{\circ}$. First, we know that $\sin(2\times67.5^{\circ})=\sin135^{\circ}=\frac{\sqrt{2}}{2}$ and $\cos(2\times67.5^{\circ})=\cos135^{\circ}=-\frac{\sqrt{2}}{2}$. Substitute into $R(\theta)=\frac{v_{0}^{2}\sqrt{2}}{32}(\sin2\theta-\cos2\theta - 1)$. $R=\frac{48^{2}\sqrt{2}}{32}(\sin135^{\circ}-\cos135^{\circ}-1)$.

Step3: Calculate the value of $R$

First, calculate the values inside the parentheses: $\sin135^{\circ}-\cos135^{\circ}-1=\frac{\sqrt{2}}{2}-(-\frac{\sqrt{2}}{2})-1=\sqrt{2}-1$. Then, $\frac{48^{2}\sqrt{2}}{32}=\frac{2304\sqrt{2}}{32}=72\sqrt{2}$. $R = 72\sqrt{2}(\sqrt{2}-1)$. Expand the expression: $R=72\times(2 - \sqrt{2})$. $R = 144-72\sqrt{2}\approx144 - 72\times1.41421\approx144 - 101.82312\approx42.18$.

Answer:

$42.18$