Projectile Motion The range R of a projectile propelled...

Projectile Motion The range $R$ of a projectile propelled downward from the top of an inclined plane at an angle $\theta$ to the inclined plane is given by $$ R(\theta)=\frac{2 v_{0}^{2} \sin \theta \cos (\theta-\phi)}{g \cos ^{2} \phi} $$ where $v_{0}$ is the initial velocity of the projectile, $\phi$ is the angle the plane makes with respect to the horizontal, and $g$ is acceleration due to gravity. (a) Show that for fixed $v_{0}$ and $\phi$ the maximum range down the incline is given by $R_{\max }=\frac{v_{0}^{2}}{g(1-\sin \phi)}$ (b) Determine the maximum range if the projectile has an initial velocity of 50 meters/second, the angle of the plane is $\phi=35^{\circ},$ and $g=9.8$ meters/second $^{2}$

Projectile Motion The range $R$ of a projectile propelled downward from the top of an inclined plane at an angle $\theta$ to the inclined plane is given by
$$
R(\theta)=\frac{2 v_{0}^{2} \sin \theta \cos (\theta-\phi)}{g \cos ^{2} \phi}
$$
where $v_{0}$ is the initial velocity of the projectile, $\phi$ is the angle the plane makes with respect to the horizontal, and $g$ is acceleration due to gravity.
(a) Show that for fixed $v_{0}$ and $\phi$ the maximum range down the incline is given by $R_{\max }=\frac{v_{0}^{2}}{g(1-\sin \phi)}$
(b) Determine the maximum range if the projectile has an initial velocity of 50 meters/second, the angle of the plane is $\phi=35^{\circ},$ and $g=9.8$ meters/second $^{2}$

Key Concepts

Projectile Motion

Projectile motion involves the motion of an object thrown into space, subject to only the acceleration due to gravity. It is characterized by a parabolic trajectory, where horizontal motion proceeds at constant velocity while vertical motion is uniformly accelerated. Understanding projectile motion requires applying kinematic equations to determine quantities such as displacement, time of flight, and range.

Inclined Plane Kinematics

Motion on an inclined plane extends the analysis of projectile motion by incorporating the geometry of the surface. It requires resolving gravitational acceleration into components parallel and perpendicular to the plane. This adjustment in the coordinate system is crucial for accurately modeling the motion of a projectile launched relative to an incline, affecting both its contact with the plane and the range along it.

Optimization in Physics

Optimization in physics typically involves using calculus to determine conditions under which a physical quantity, such as range or energy, is maximized or minimized. In projectile motion, finding the optimal launch angle to maximize range often requires differentiating the range equation with respect to the launch angle and setting the derivative equal to zero to solve for the extremum.

Trigonometric Relationships in Kinematics

Trigonometric relationships are essential in projectile motion problems, especially when dealing with inclined planes and non-standard launch angles. Manipulating expressions involving sine, cosine, and their combinations is necessary to simplify equations, derive formulas, and ultimately solve for the desired quantities such as maximum range or optimal angles.

Transcript

00:01 So in this problem, we're going to simplify this formula for range given by r to find the maximum r.

00:09 And we know what it's meant to be.

00:11 We just have to take some steps using our formulas and identities to get there.

00:15 And then in part b, we'll plug in some values for v -0, g, and fee to find a specific range for a given set of circumstances.

00:25 So a formula for r is given by this, and we need to transform it a little bit.

00:30 So let's start by transforming the numerator using a product to sum formula, since it's a product of a sign and a cosine.

00:39 So we have two times v -not squared, and using our formula, we get one -half times sign of the sum, so theta plus theta -fee, plus sign of the difference.

00:55 So theta -minus theta -v.

00:59 And we won't do anything with the denominator just yet.

01:05 And simplify the numerator.

01:07 The two and the one half will cancel each other out.

01:10 So we have v0 squared out here.

01:13 And then we'll simplify within these sign functions.

01:16 So the argument of the first sign function will be two theta minus fee.

01:23 And for the second one, the thetas will cancel out and our negatives will cancel out.

01:28 So we'll have sign of fee for the second.

01:35 And we'll that we've simplified this a bit, let's start thinking about the maximum this could be.

01:42 So fee is fixed, so we don't know how big fee is going to be, but theta will vary.

01:48 And so what is really important here is that sign of something can never be larger than one...

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