/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 81 A world-class volleyball player ... [FREE SOLUTION] | 91Ó°ÊÓ

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A world-class volleyball player can jump vertically \(1.1 \mathrm{~m}\) from a standing start. (a) How long is the player in the air? (b) Graph the athlete's position versus time. (c) Use your graph to explain why an athlete might appear to "hang" in the air near the top of the jump.

Short Answer

Expert verified
The player is in the air for about 0.946 seconds. The position vs. time graph shows a parabola with a slow change near the peak, creating a 'hang-time' illusion.

Step by step solution

01

Understanding the Jump's Peak Height

The player's jump reaches a peak height of \(1.1 \mathrm{~m}\). To find the time in the air, we need to understand the physics of vertical motion under gravity alone. The time to reach the peak is the same as the time to fall back to the original jump position.
02

Using Kinematic Formula to Find Time to Peak

The kinematic formula for vertical motion from rest is given by \[ v^2 = u^2 + 2as \]where \( v = 0 \mathrm{~m/s} \) at the peak (since the player momentarily stops before falling), \( u \) is the initial velocity, \( a = -9.81 \mathrm{~m/s^2} \) (acceleration due to gravity), and \( s = 1.1 \mathrm{~m}\). Rearranging to find \( u \):\[ 0 = u^2 - 2 \times 9.81 \times 1.1 \]\[ u^2 = 21.582 \]\[ u = \sqrt{21.582} \approx 4.64 \mathrm{~m/s} \].
03

Calculating Time to Reach Peak

Use the formula \[ v = u + at \]Substituting \( v = 0 \), \( u = 4.64 \mathrm{~m/s} \), and \( a = -9.81 \mathrm{~m/s^2} \): \[ 0 = 4.64 - 9.81 \times t \]\[ t = \frac{4.64}{9.81} \approx 0.473 \mathrm{~s} \].
04

Total Time in Air

Since the time to reach the peak is \(0.473 \mathrm{~s}\), the total time for the complete jump (up and down) is double that time:\[ t_{total} = 2 \times 0.473 \approx 0.946 \mathrm{~s} \].
05

Graph Position vs. Time

To graph position versus time, plot a parabola, where the vertex represents the peak of the jump at \(0.473 \mathrm{~s}\) and \(1.1 \mathrm{~m}\). The slope is steepest at the start and end, becoming flat at the peak.
06

Explaining the Hang-time Appearance

The athlete seems to 'hang' at the peak because the velocity is very small (near zero) at the top of the jump. This makes the change in vertical position small and slow compared to other times during the jump, giving the illusion of a longer time spent at the peak.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinematics and Projectile Motion
Kinematics is a branch of physics that deals with the motion of objects without considering the causes of motion. It’s all about describing how objects move in terms of displacement, velocity, and acceleration. In the context of projectile motion, which involves objects being launched into the air, kinematics helps us understand the path that these objects take.

When a volleyball player jumps, their motion can be broken down into two dimensions: horizontal and vertical. For a vertical jump, like in this exercise, kinematics mainly involves vertically upward motion followed by free fall. The critical concept here is that the only force acting on the player is gravity, which affects the vertical motion.

Key equations in kinematics help us solve for various quantities associated with motion:
  • The equation: \[ v^2 = u^2 + 2as \]This helps determine the change in velocity of the player as they ascend and descend.
  • The equation: \[ v = u + at \]This one helps calculate the time it takes for the player to reach the highest point in their jump.
Understanding these principles allows us to model the player’s jump accurately.
Vertical Motion in a Jump
Vertical motion refers to any movement that happens along the vertical plane. In this volleyball player's jump, the motion began with an upward thrust and ended with a descent back to the ground level.

It's essential to understand that during the ascent and descent, the athlete's body undergoes acceleration and deceleration. However, the acceleration is constant during the entire motion due to gravity, which is approximately \(9.81 \, \text{m/s}^2\) downward.

In reaching the jump's peak, the initial velocity propels the player upwards, eventually reaching a momentary stop before gravity pulls them back down. This dynamic is captured within the kinematic equations and helps explain how long the player stays airborne and how high they jump.

One of the surprises for many is that the time taken to ascend is equal to the time taken to descend. This symmetry is due to the constant gravitational force that acts equally on ascent and descent. By calculating just the ascent time using the formula \[ t = \frac{u}{g} \], we can double it to find the total time the player spends in the air.
Understanding Free Fall
Free fall is a type of vertical motion where gravity is the only force acting on an object. During the free fall phase of any vertical jump, like our volleyball player's descent, the object accelerates downward due to gravity, experiencing constant acceleration.

At the jump’s peak, the concept of free fall becomes most apparent. Here, the initial upward velocity is zero, and gravity takes over entirely. Despite being termed 'free fall', the object might still be moving upwards. Then, as the velocity turns to zero and reverses, the object accelerates downward until it hits the ground.

The peculiarity of free fall, and why it can appear as a ‘hang’ time for the athlete, lies in the small changes in velocity at the jump's peak. As the velocity reduces to nearly zero before reversing, the motion appears to slow, giving an optical illusion that the athlete is suspended mid-air.

Knowing about free fall helps us understand how long objects are in the air and their position over time. It explains why athletes may seem to hang in the air and why this phase dominates the viewer's perception of the jump duration.

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