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In the vertical jump, an athlete starts from a crouch and jumps upward as high as possible. Even the best athletes spend little more than 1.00 s in the air (their "hang time"). Treat the athlete as a particle and let \(y_{max}\) be his maximum height above the floor. To explain why he seems to hang in the air, calculate the ratio of the time he is above \(y_{max}\)/2 to the time it takes him to go from the floor to that height. Ignore air resistance

Short Answer

Expert verified
The ratio is 1.

Step by step solution

01

Identify Given Information and Define Variables

We are given that the total time in the air (hang time) is 1.00 s. We need to calculate the ratio of the time the athlete is above half of his maximum height, \( \frac{y_{max}}{2} \), to the time taken to reach that height from the floor. Let's define time to maximum height as \( t_1 \) and time above half maximum height as \( t_2 \).
02

Use Symmetry in Projectile Motion

In projectile motion without air resistance, the time taken to reach maximum height (\( t_1 \)) is half of the total hang time. Thus, \( t_1 = \frac{1.00}{2} \ \text{s} = 0.50 \ \text{s} \). We need to find how much time the athlete spends above \( \frac{y_{max}}{2} \).
03

Apply Kinematic Equations

From the equation \( y = v_i t + \frac{1}{2} a t^2 \), since the initial velocity (\( v_i \)) at \( y_{max} \) is 0, and using acceleration due to gravity \( a = -9.8 \ \text{m/s}^2 \), we have:\[ 0 = v_0 t_1 - \frac{1}{2} g t_1^2 \]Solving for \( v_0 \), we find \( v_0 = g t_1 \). This establishes the relation between initial velocity and the time to reach maximum height.
04

Determine Time to Reach Half Maximum Height

By symmetry, it takes half the time to fall from \( y_{max} \) to \( \frac{y_{max}}{2} \) as it does to fall from \( \frac{y_{max}}{2} \) to the floor. Given this, the time from maximum height to \( \frac{y_{max}}{2} \) is \( \frac{t_1}{2} \).
05

Calculate Time Above Half Maximum Height

The time above \( \frac{y_{max}}{2} \) includes both the time descending from maximum height to \( \frac{y_{max}}{2} \) and the same time ascending from \( \frac{y_{max}}{2} \) to the maximum height due to symmetry in projectile motion. Thus, \( t_2 = t_1 - \frac{t_1}{2} \).
06

Calculate the Desired Ratio

Using all previous steps, the time the athlete spends above \( \frac{y_{max}}{2} \) is \( t_2 = 0.5 - 0.25 = 0.25 \ \text{s} \). The time to climb from floor to \( \frac{y_{max}}{2} \) is \( \frac{t_1}{2} = 0.25 \ \text{s} \). Hence the ratio is:\[ \text{Ratio} = \frac{0.25}{0.25} = 1 \]
07

Conclusion

The ratio of the time the athlete is above \( \frac{y_{max}}{2} \) to the time it takes him to go from the floor to that height is 1.

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Key Concepts

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

Kinematic Equations
Kinematic equations are fundamental when analyzing motion, particularly under uniform acceleration like in projectile motion. For any projectile, these equations typically relate four key variables: displacement, initial velocity, time, and acceleration.
  • The basic kinematic equation used in projectile motion is: \[ y = v_i t + \frac{1}{2} a t^2 \]where \(y\) is the displacement, \(v_i\) is initial velocity, \(t\) is time, and \(a\) is the constant acceleration due to gravity (\(-9.8 \ \text{m/s}^2\)).
  • Since gravity is a downward force, the acceleration component \(a\) is negative.
In the exercise example, these equations help determine initial velocity and timings to predict the motion path of the jumping athlete. The use of kinematic equations provides a structured approach to solve complex problems by dissecting them into simpler, calculable segments. By knowing at least three of these key variables, the other one can be resolved easily, making it easier for students to understand and predict the movement of projectiles.
Symmetry in Motion
Symmetry is a crucial aspect of projectile motion that simplifies calculation. When an object is launched upward and then descends back to the ground, the motion can be thought of as two mirror halves.
  • The time taken to ascend to the peak of its flight is equal to the time taken to descend back from the peak to the initial launch height.
  • This makes the total flight time twice the time it takes to reach the maximum height.
In our solution, we highlight this symmetry by identifying that the hang time (1.00 seconds) is equally split between the ascent and descent, giving an ascent time \(t_1 = 0.50 \ \text{s}\). Symmetry also implies equal time on either side of any horizontal line cutting the path of the jump evenly (e.g., half maximum height). This property makes determining time segments between height points straightforward, drastically reducing complexity in calculations and visualizing motion paths.
Vertical Jump Analysis
Analyzing a vertical jump using physics concepts like projectile motion provides insight into the dynamics of the jump. A vertical jump is essentially a textbook example of a projectile launched vertically.
  • The athlete's vertical jump can be equated to a projectile's trajectory, traveling upwards to a maximum height before gravity pulls it back down.
  • Hang time or total time in the air, primarily depends on the initial velocity of the jump. Still, the distribution of time at specific heights is symmetric.
In the exercise, the jump is analyzed to understand why the athlete seems to have additional hang time as they reach the peak. By calculating the time spent above half the maximum height, the results demonstrate this apparent hang due to the prolonged time spent near the vertex of the projectile's arch. The practical application involves applying kinematic equations and appreciating motion symmetry, enabling predictions about the athlete's physical capabilities and jump dynamics. Hence, physics techniques offer a great understanding of not only everyday motions but also athletic performances.

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