A basketball star covers 2.80 m horizontally in a jump to...

A basketball star covers $2.80 \mathrm{~m}$ horizontally in a jump to dunk the ball (Fig. P4.23). His motion through space can be modeled as that of a particle at a point called his center of mass (which we shall define in Chapter 9). His center of mass is at elevation $1.02 \mathrm{~m}$ when he leaves the floor. It reaches a maximum height of $1.85 \mathrm{~m}$ above the floor and is at elevation $0.900 \mathrm{~m}$ when he touches down again. Determine (a) his time of flight (his "hang time" , (b) his horizontal and (c) vertical velocity components at the instant of takeoff, and (d) his takeoff angle. (e) For comparison, determine the hang time of a whitetail deer making a jump with center-of-mass elevations $y_{i}=1.20 \mathrm{~m}, y_{\max }=2.50 \mathrm{~m}, y_{f}=0.700 \mathrm{~m}$.

Key Concepts

Projectile Motion

This is the study of the motion of objects that are launched into the air and subject only to the acceleration due to gravity. The motion is typically decomposed into independent horizontal and vertical components, where the horizontal motion has a constant velocity and the vertical motion is uniformly accelerated. Understanding projectile motion is essential for analyzing trajectories in various real-life contexts.

Kinematic Equations

These equations describe the motion of a particle under constant acceleration. They are crucial in relating displacement, initial velocity, time, and acceleration. In projectile motion problems, different kinematic equations are applied separately to the horizontal and vertical components to determine factors such as time of flight, maximum height, and velocity components.

Decomposition of Vector Components

This concept involves breaking down an initial velocity vector into horizontal and vertical components. The horizontal component is responsible for covering distance over time without acceleration (ignoring air resistance), while the vertical component, influenced by gravity, determines the change in elevation and the time it takes for the projectile to reach its maximum height and descend.

Time of Flight

Time of flight, often referred to as hang time in the context of projectile motion, is the total time an object spends in the air from launch to landing. It is determined by analyzing the vertical motion of the projectile, utilizing the appropriate kinematic equations that account for initial vertical position, maximum height, and landing height.

Initial Velocity and Launch Angle

The magnitude of the initial velocity and its direction (launch angle) are central to understanding the projectile's motion. The initial velocity is resolved into horizontal and vertical components using trigonometric functions, and the launch angle connects these components. Accurate determination of these values is essential for predicting the trajectory and range of a projectile.

Transcript

00:01 Hello.

00:02 In this problem, we have a basketball player that can jump using his center of mass, which is a concept that you will get to know later on moving in this book.

00:16 So his center of mass was initially at height 1 .02 meters, so that's the initial.

00:24 So why initial? 1 .02 meters.

00:31 And the maximum height that he reaches when he is jumping is 1 .85 meters, so that's why maximum.

00:46 And when he get back to touch the earth again, he's not at the same height that he was initially, but actually a little bit lower and he stands at 0 .9 meters.

00:59 So let's call this the y final.

01:02 The player covers 2 .8 meters in distance, in his jump, and we're asked to determine few things about his motion.

01:17 So let's start first by getting what we can get using only these givings.

01:25 So from these givings, we can actually determine the velocity, or at least the vertical component of his velocity.

01:32 Using the formula of v i y squared v f y squared is equal to v i y squared plus 2 g y this is from kinematics and in order to get the time that it takes or to make like our parameters a little bit less we use that point at the apex which is the maximum height so that his vertical velocity at this point is typically zero.

02:13 And now we will have that his initial velocity squared is equal to negative to gy, and here g is already negative because it's pointing downward.

02:24 So that's equal to 19 .6y.

02:29 And y here, be careful.

02:31 Y here is the distance.

02:33 He moved from his initial position to the final position.

02:36 So it's better to denote this as delta y.

02:41 And delta y here is just has maximum position, subtracted from the initial position.

02:52 And doing this on your calculator, you will know that this is equal to 16 .268.

03:00 So v .i .y taking the square root, it's 4 .033 meters per second.

03:08 And that's the vertical component of the player's velocity.

03:13 And actually, this is the answer for part c.

03:17 You usually can start with whatever you find easier to get, regardless of the order in the problem itself.

03:26 So that's the first thing we got here, the vertical component of the velocity.

03:30 And using this, we can actually determine the total time that he kept hanging in the air.

03:37 Or it's called the hang time in short.

03:40 So let's start by using the second equation of kinematics, which is delta y is equal to vii, t minus half multiplied by 9 .8 t squared.

04:00 Here, delta y, we will use because we are interested in the total hang time, so here we will be interested in the initial height of his center of mass, which is 1 .02, and the final height, which is 0 .9.

04:15 So here, delta y is basically 0 .9 minus 1 .02.

04:26 And v -i -y, as we got in the previous part, it's 4 .033 -t minus 4 .9 t squared.

04:39 Doing this on your calculator, this is negative 0 .12.

04:43 So let's replace this with negative 0 .12.

04:50 You can re -advenge this or block it on your calculator as it is to find out that we have two roots for this quadratic equation.

05:01 And the first one is t equal to 0 .852...

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