REFERRING TO EXAMPLE 4-11 Suppose the ball is dropped at the horizontal distance of 5.50 m, but

step 1 Question 94 deals with example 411 in the book on page 102. So I recommend looking at that first

REFERRING TO EXAMPLE 4-11 Suppose the ball is dropped at the...

Key Concepts

Free Fall and Gravitational Acceleration

This concept describes the motion of objects under the sole influence of gravity, which accelerates any object downward at about 9.81 m/s². Understanding free fall is key to analyzing the vertical motion component in projectile problems.

Relative Timing in Motion

Determining when and where moving objects intersect involves setting up equations so that their positions are equal at the same time. This concept is fundamental in scenarios where two moving bodies must meet, such as a dolphin catching a ball in mid-air.

Launch Angle

The launch angle is the angle at which a projectile is released relative to the horizontal. This parameter directly affects the distribution of the initial velocity into horizontal and vertical components, which in turn influences the projectile's overall trajectory and range.

Minimum Speed Determination

In problems requiring the calculation of the least initial speed needed to guarantee a successful catch or clearance, optimization methods are used. This involves solving the motion equations under specific constraints to ensure that all conditions, such as meeting a moving target before an obstruction, are satisfied.

Kinematics Equations

Kinematics involves using equations that relate displacement, initial velocity, acceleration, and time for objects in motion. These equations are essential for determining positions and velocities at different points in time for projectiles and objects in free fall.

Projectile Motion

This concept examines the two-dimensional motion experienced by an object that is launched, where its horizontal component moves at constant velocity while the vertical component accelerates due to gravity. Analyzing projectile motion involves understanding how these independent components combine to form the object's parabolic trajectory.

Decomposition of Motion

This involves breaking the initial velocity vector into horizontal and vertical components using trigonometric functions. Such decomposition is crucial for separately analyzing the motion along each axis, thereby determining factors like range, flight time, and maximum height.

Transcript

00:01 Question 94 deals with example 411 in the book on page 102.

00:07 So i recommend looking at that first before going to this problem, just to give a little more context about what the problem is about.

00:17 So assuming you've done that already, we'll move on to the solution to this question in which it states, supposedly ball has dropped at a horizontal distance of 5 .5 meters away from a dolphin, but in this scenario a new height of 5 meters.

00:33 The dolphin jumps with the same speed of 12 meters per second.

00:36 Part a of the question asks, what launch angle must the dolphin have if it is to catch the ball? so there's three parts.

00:42 We'll go part by part here.

00:44 So the first part here asks, what is theta here if it wants to catch the ball? so the dolphin is at the bottom left corner here.

00:52 It's aiming for the ball.

00:54 It will launch towards the direction of the ball, but due to gravity, and of course, will start to curb down, in which the two, the dolphin and the ball, will meet at some point further along.

01:03 So in order for this to happen, the angle of theta here, we just blow up this triangle a little bit, this here inside is theta, this is h and this is d.

01:14 Let's pretend this is a perfect right triangle, because i pour draw or sometimes.

01:18 We know that the tangent of the angle here is opposite or adjacent, so the arc tan of theta is simply the height over the distance.

01:30 These variables again, these were chosen in the example, so i'm just sticking with them.

01:34 Typically i'd use x and y, but we'll go it this way.

01:38 So it's the height over the distance, so that's the 10 inverse of 5 over 5 and a half, which results in an angle to three significant figures of 42 .3 degrees.

01:48 So that's a solution for part a.

01:52 Part b now, we read, says at what height does the dolphin catch the ball in this case? so looking for this height here, between here and here, note that my picture may not be to scale.

02:06 I want to call this height d, y.

02:08 So if i use y, i should indicate some directions, x and y.

02:14 I'm going to call this d .y.

02:16 So if i know based on what i'm given, v, naught, and acceleration and such, i'm going to start with this equation.

02:22 So my distance y will be positive because i'm going from a position of lower to higher, equals my initial velocity in the y direction times time, minus because the acceleration due to gravity is negative, minus one -half g -t squared.

02:42 So yeah, so i'm here.

02:44 I'm at this point.

02:44 I recognize i have my initial velocity.

02:47 I have d -y.

02:48 I have g, of course, but i don't have time.

02:52 So i'm going to look back to my x equation for the scenario.

02:55 So my distance d is equal to the initial velocity in my x direction.

03:02 Times time, which i can solve for that to be distance, d as given in the figure, remember, over my initial velocity times cosine of the angle.

03:18 So i have this solution, i know all the terms, so i can solve for numerically for t.

03:22 However, i do like to keep things in terms of variables, so that's what i will do, so i don't get lost in numbers.

03:29 So i'm going to substitute this t into my equation here, and therefore i'll have every term except for d, y, the one i'm solving for.

03:35 So my initial velocity in the y direction is v .0 sine theta.

03:40 Multiply that by d over v .0 cosine theta, because that's my time term minus g over 2 times this t term squared, which is this d squared over v .0 squared cosine squared data.

04:04 So just a quick cleanup, i have my v0, sorry, my v0 terms will cancel out.

04:09 And left from the expression d tan theta minus g d squared over to v.

04:24 Not squared cosine squared theta...

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