Projectile Motion An object is propelled upward at an angle θ, 45^∘<θ<90^∘, to the horizontal wi

Projectile Motion An object is propelled upward at an angle...

Projectile Motion An object is propelled upward at an angle $\theta, 45^{\circ}<\theta<90^{\circ},$ to the horizontal with an initial velocity of $v_{0}$ feet per second from the base of a plane that makes an angle of $45^{\circ}$ with the horizontal. See the figure atop the right column. If air resistance is ignored, the distance $R$ that it travels up the inclined plane is given by the function $R(\theta)=\frac{v_{0}^{2} \sqrt{2}}{16} \cos \theta(\sin \theta-\cos \theta)$(a) Show that$$ R(\theta)=\frac{v_{0}^{2} \sqrt{2}}{32}[\sin (2 \theta)-\cos (2 \theta)-1] $$ (b) In calculus, you will be asked to find the angle $\theta$ that maximizes $R$ by solving the equation $$ \sin (2 \theta)+\cos (2 \theta)=0 $$ Solve this equation for $\theta .$ (c) What is the maximum distance $R$ if $v_{0}=32$ feet per second? (d) Graph $R=R(\theta), 45^{\circ} \leq \theta \leq 90^{\circ}$, and find the angle $\theta$ that maximizes the distance $R .$ Also find the maximum distance. Use $v_{0}=32$ feet per second. Compare the results with the answers found in parts (b) and (c).

Key Concepts

Graphing Functions

Graphing functions is a powerful method for visualizing how a function behaves over a range of inputs, verifying analytical solutions, and identifying key features such as maximum or minimum points. By graphing the range function with respect to the launch angle, one can confirm the optimal angle found through calculus and gain deeper insight into the relationship between the launch angle and the distance traveled along an inclined plane.

Calculus-based Optimization

Calculus-based optimization involves using derivatives to identify the maximum or minimum values of a function. In the context of projectile motion, taking the derivative of the range function with respect to the launch angle and setting it to zero can help find the angle that maximizes the distance traveled. This technique is widely applied in physics and engineering to optimize performance or behavior based on given parameters.

Projectile Motion

Projectile motion is the study of objects launched into the air, where the only significant force acting on them is gravity (assuming air resistance is negligible). This concept examines the trajectory, range, and time of flight of the projectile. In problems involving inclined surfaces, the motion analysis is adjusted to account for the plane’s orientation, which can alter the effective components of the motion.

Trigonometric Identities

Trigonometric identities, such as the double-angle formulas, are crucial tools for rewriting and simplifying expressions involving trigonometric functions. These identities allow for the transformation of equations into forms that are easier to analyze and solve, especially when the problem requires comparing different trigonometric expressions or finding maximum values through algebraic manipulation.

Transcript

00:01 In this problem, we're given function r, which represents the distance and object travels up the inclined plane, as shown in this diagram.

00:11 So part a is asking us to show that these two functions are actually equal to each other.

00:17 So therefore, what we're going to do is we're going to start with the giving information, and we're going to distribute the expression in front of parentheses to get rid of it.

00:29 So therefore, this is the same thing as initial velocity square, square root over 16 times cosine of theta and sine of theta minus initial velocity square root over 16 of cosine square theta.

00:48 So now if you remember that sine of two theta double angle formula is same thing as two, sine of theta times cosine of theta.

00:59 So if i divide it by 2, then that's going to give us what cosine of theta and sine of theta equals 2.

01:06 So therefore, this part i can rewrite it as initial velocity square root 2 over 16 times.

01:15 So sine of theta times cosine of theta is same thing as sine of 2 theta over 2.

01:23 And then minus initial velocity square root over 16.

01:28 And we know what cosine square theta is, which is one of the half angle formula, that's equivalent to 1 plus cosine of 2 theta over 2.

01:41 So now i'm going to get rid of this 2.

01:44 So this is the same thing as 1 half times 1 over 16.

01:47 So therefore we give initial velocity square square 2 over 32 times sine of theta.

01:56 Minus same thing i'm going to do one half times one over 16 so we have initial velocity square square of two over 32 times one plus cosine of theta two theta so now i can see that both terms have initial velocity square square two over 32 so i'm going to factor that out so what we have left is sine of theta sign of i'm sorry i forgot 2 theta here, sine of 2 theta, and then minus 1 plus cosine of 2 theta.

02:43 And now what we're going to do is we're going to simplify what's inside the parentheses.

02:48 So as you can see, we have demonstrated that the function r is equivalent to initial velocity square square root 2 over 32 times sine of 2 theta minus code.

03:05 Cosine of 2 theta minus 1.

03:08 And that's exactly what we were trying to prove right here.

03:11 Okay, exactly the same thing.

03:16 Sign of 2 theta minus 2 cosine.

03:19 So it's exactly the same thing.

03:20 So now next part, b, b is asking us to solve this equation, sine of 2 theta plus cosine of 2 theta equals to 0.

03:37 Because solving for theta will help us find the angle theta that maximizes that distance travel.

03:46 So therefore, the first thing i'm going to do is i'm going to subtract cosine of 2 theta on both sides.

03:52 So therefore, we have sine of 2 theta equals to negative cosine of 2 theta.

04:00 And now i'm going to divide both sides by cosine of 2 theta.

04:03 We know by definition, sine over cosine is same thing as tangent.

04:12 So we have tangent of 2 theta equals to negative 1.

04:17 And now what i'm going to do is i'm going to let u equals to 2 theta.

04:23 Okay, so therefore i'm looking for tangent of u equals to negative 1.

04:31 So you're going to look at the unit circle.

04:34 So in order to get negative 1, the x and the y is going to have same number where one is negative and one is positive.

04:43 So therefore, the two possible solution is when u is at 3 pi over 4, okay, square 2 over 2 divided by negative square 2 over 2, it's going to be negative 1.

04:55 And then the other one is located at 7 pi over 4.

04:59 Same thing.

04:59 It's going to give you negative 1 as well when you look at the cosine and sine value.

05:05 So therefore, we know that u equals to 3 pi over 4 and 7 pi over 4.

05:19 So now we know what u equals to.

05:22 So now i can substitute u into this equation.

05:25 So this is 3 pi over 4 equals to 2 theta.

05:30 So when you divide it by 2, so theta is going to be equal to 3 pi over 8.

05:37 And then next one, i'm going to let you equals to 7 pi over 4 equals to 2 theta again divided by 2.

05:49 So theta is going to be equal to 7 pi over 8.

05:55 Now remember we have a domain restriction for this.

06:00 The domain restriction for this is that theta has to be between 45.

06:08 And 90 degrees.

06:13 So it has to between 45 degrees, between 45 and 90 degrees, okay? so therefore you can multiply this by 180 over pi to see how many degrees that is.

06:26 And when you do that in your calculator, you get about 67 .5 degrees.

06:33 And next one, same thing, multiply this by 180 over pi.

06:38 So theta in degrees is about about 157 .5 degrees.

06:44 So the only acceptable answer is that theta has to be 3 pi over 8.

06:50 That's the answer.

06:51 This one can be it, okay? because it's not in the domain restriction.

06:57 So now we know the theta.

06:59 So question c, part c is asking you to find the maximum distance if the initial velocity is 32 feet per.

07:11 Second.

07:12 So we know that from the from early equation we know that the function is equal to initial velocity square square root to over over 32 times sine of two theta minus cosine of two theta minus one.

07:38 So we're gonna plugging our theta so our theta here is is 3 pi over 8.

07:48 And they gave you the initial velocity, which is 32...

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