The meadow jumping mouse, Zapus hudsonius, inhabits much of North America and can jump as high a

step 1 So we're doing an example with projectile motion, reminder. Projectile motion is a particular fo

The meadow jumping mouse, Zapus hudsonius, inhabits much of...

Key Concepts

Trigonometric Relationships

Trigonometric functions, such as sine and cosine, appear when resolving the initial velocity into its components. Their relationships are crucial for linking the magnitude of the initial velocity with its directional components, which in turn are used to satisfy the geometric constraints of the problem, such as clearing an obstacle at a specific distance.

Trajectory Optimization to Meet Constraints

This involves determining the initial conditions (such as speed and angle) required for a projectile to reach or just clear a specified obstacle. It requires setting up equations that satisfy the condition at a particular point (for example, the obstacle’s height at a given horizontal distance) and solving for the minimal or optimal parameters satisfying that criterion.

Kinematic Equations

These equations describe the position, velocity, and acceleration of an object over time. In projectile problems, they are used to relate the horizontal displacement with time (using constant velocity) and the vertical displacement with time (using constant acceleration due to gravity), which together define the trajectory of the projectile.

Decomposition of Velocity into Components

In projectile motion, the initial velocity is resolved into horizontal and vertical components. The horizontal component (v*cos?) is constant throughout the flight (ignoring air resistance), while the vertical component (v*sin?) is affected by gravity. This separation is essential for analyzing the motion independently in each direction.

Projectile Motion

This concept deals with the motion of an object that is launched into the air and is subject to gravitational acceleration. It involves analyzing the object's trajectory, which is parabolic in the absence of air resistance, and requires the use of kinematic equations to relate the object's initial velocity, launch angle, and displacement over time.

Transcript

00:01 So we're doing an example with projectile motion, reminder.

00:08 Projectile motion is a particular form of motion where your ex component parallel to the earth remains uh constant velocity.

00:25 So if you start with some initial velocity, continue with that velocity.

00:30 And it was starting at the origin.

00:33 Um the x motion just builds up as that velocity times time.

00:39 Great times time equals distance.

00:42 Um the wind motion.

00:43 If we're talking about gravity, what happens is your your y component of velocity changes in time and just gets smaller or more negative due to gravity.

01:01 Uh and that makes the y motion.

01:04 If we're starting at the origin a little bit more complicated than rate times, time equals distance.

01:11 There's a minus one half gt squared in that motion as well.

01:18 Um and one class of problems that deals with projectile motion are what are called what i'll call clearing the fence type problems where you have some sort of obstacle, it may be a tennis net.

01:34 Um it could be the home run wall out baseball and the idea is you want your projectile to get over that obstacle.

01:46 Um so as an example of this type of problem, let's suppose we have some critter that wants to get into your garden and eat your goodies.

01:55 Um that's a critter.

01:58 Uh we won't worry about what it is, but it has a tail, so it's not just a point.

02:04 Um so your your garden wall will say is 48.3 cm high, which you think is high enough to stop most critters.

02:16 Um and that critter is um 62 cm away uh safely far enough from you.

02:28 And the question is how fast and in what direction does that critter have to jump? so what kind of show a v.

02:38 Zero vector velocity vector at an angle? and the question is how fast which is vienna magnitude of that speed? and in what direction does the little creature have to jump to just barely get over that wall? um just barely clear it.

03:02 Um so we can solve this with our projectile motion equations.

03:08 The idea of just barely making it all right, tells us that at the point in the little critters motion where it's it's getting over um it only has uh it's vi zero x.

03:30 Component a velocity which isn't changing and there's no why component of velocity.

03:43 So it's at the height of its, its its motion.

03:47 Sorry, that should be b.

03:48 Y.

03:49 Equals zero.

03:51 Um and so that's what we have to work with is two different dimensions to begin with.

03:58 And the point where the creatures just clearing the fence.

04:02 Um and a lot of students want to set the tangent of the initial angle equal to the 48.3 cm over the 62 cm.

04:16 And that's not quite correct because um the path is going to curve to the gravity.

04:23 Um you'll follow a nice parabolic path.

04:27 And so when the creature gets to the top of that that wall um the angle is going to flatten out.

04:35 So obviously it's going to have to jump with a much steeper angle than 48.3 over 62 being the tangent.

04:44 Um and that's one of the things that we're going to have to solve for.

04:48 So we are solving for the v.

04:50 Not and the angle um to just barely clear the fence.

04:56 Uh so let's look at what this implies for the equations.

05:00 Um obviously we've already addressed that the v.

05:04 Y.

05:04 Has to be zero which means that the zero y.

05:10 Has to equal g.

05:11 T.

05:13 And that's going to fix the time.

05:15 Now.

05:16 We we really don't care how much time this little guy takes to get over.

05:19 Um but what we can do is substitute t.

05:25 With v.

05:26 Not y over g.

05:30 And flood reminder is that r.

05:33 V.

05:34 Not why we should write down our components very clearly rv not why is the same as the magnitude of the speed or the magnitude of the initial velocity times the sine of the initial angle.

05:50 And meanwhile we have a similar thing for vi0 x.

05:55 Is the magnitude of the initial velocity times the co sign of the initial angle.

06:05 Okay.

06:06 And that's what we're going to be working with, is uh now getting out of our nice clean component form and getting our equations in terms of these two unknowns.

06:19 Um and we're gonna wind up with two equations into unknowns.

06:24 Um we're going to be working with the x.

06:28 Equation and the y.

06:29 Equation for which we know the distances.

06:33 Um and we're going to be substituting in for the two unknowns at the b0 and the unknown angle.

06:42 So let's kind of just see how those equations work out if we make our substitution in time.

06:52 Uh huh.

06:53 And we're going to substitute this in two different places in the uae equation and the x.

06:58 Equation for which we have the most information and that will give us two new equations.

07:12 Let's see if we could scroll down on our board.

07:18 There we go.

07:19 Um so let's go ahead and substitute goes in the x.