Rescue Swimmers Coast Guard rescue swimmers are trained to...

Key Concepts

Projectile Motion

Projectile motion describes the motion of an object that is launched into the air and moves under the influence of gravity alone, without propulsion. This type of motion is characterized by a parabolic trajectory, where the horizontal and vertical components of the motion are independent of each other except for the common time parameter.

Vector Decomposition

Vector decomposition is the process of breaking a velocity vector into its horizontal and vertical components. This is crucial in analyzing any motion involving two dimensions, as it allows each component to be studied separately; for example, the horizontal component remains constant while the vertical component is affected by gravitational acceleration.

Gravitational Acceleration and Free Fall

Gravitational acceleration is the constant acceleration due to gravity acting on objects near the Earth’s surface. In the context of free fall, an object that starts with an initial vertical velocity will experience a constant acceleration downward, leading to a change in its speed and the curvature of its motion path.

Kinematic Equations

Kinematic equations are used to relate the initial velocity, acceleration, displacement, and time in situations of uniform acceleration. These equations are fundamental in predicting the behavior of objects in free fall and in determining how long it takes for an object to travel a certain vertical distance.

Transcript

00:01 All right, questions 70 in the book states that coast guard rescue swimmers are trained to leap from helicopters into the sea to save boaters in distress.

00:10 The rescuers like to step off their helicopter when it is, quote, 10 and 10, which means that it is 10 feet above the water and moving forward horizontally at 10 knots.

00:21 The questions are for a, what are the speed and b, the direction of motion as the rescue swimmer enters the water following a 10 and 10 jump? this is an interesting problem.

00:33 So we have a helicopter, which i can only draw as a triangle, a rectangle.

00:37 I can't.

00:38 Not a good draw.

00:38 So once the helicopter is 10 feet above the water and moving horizontally with vx at 10 knots is this 10 -and -10 situation called.

00:52 So because i use x, let me first draw a cartesian corridor system before i move on.

00:57 So our initial velocity vx is given to us in our d -y.

01:02 Our distance above the water.

01:05 So first thing we should do is convert these both to mks units.

01:09 So for 10 feet, sorry, in one foot there's 3 .28, sorry, for one meter, that's 3 .28 feet.

01:17 So if we can convert this, we have 3 .05 meters.

01:23 That one i knew, this one i did not know.

01:24 I had to look up.

01:26 So for one knot, that corresponds to 0 .514 meters per second.

01:38 So the helicopter is traveling at 5 .14 meters per second, of course.

01:48 Great, so we have our two parameters that we're given to us in our proper units, and we can go from there.

01:55 So our end goal, we want to solve for the velocity and time as the rescue swimmer enters the water, something like this, where if you blow up that final situation, you know that there's the vx component of that velocity.

02:09 So really, this should be v0 .x, maybe, v.

02:16 Not our initial velocity, where the final velocity, upon entry would be vx and vy at some angle below the horizontal.

02:29 So generally with this scenario here.

02:33 Okay, so we know we need to find vx and vy.

02:35 The final velocity is in each instance.

02:38 So the first thing we want to find is the velocity in the y direction.

02:42 So we can use an equation like this where we have our vy is v0y plus a yt.

02:56 So note here, while our initial velocity is zero, because we're moving horizontally, off the helicopter.

03:04 So now we're left with our final velocity in the y direction is g times t however we don't know what t is as it right now.

03:13 So we need to find the give yourselves an equation for time.

03:20 So if you look at one of our other distance or y equations which i'll start down here we have v0 y t plus one half a yt square.

03:39 So again, our initial velocity in the y direction is zero, and we can obtain a value for t here.

03:46 If we note now, just thinking about our signage, if we're going from a place of higher height to lower height, because we define y to be upward, but to be positive, we would have a negative term in front of each of these terms...

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