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Chapter 2: Problem 87

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 \(1.00 \mathrm{~s}\) 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
To calculate the ratio, first, find the time it takes to ascend to \(y_{\max}/2\) from the floor (\(t_{1/2}\)). Then find the time spent above \(y_{\max}/2\) by subtracting \(t_{1/2}\) from \(t_{\max}\) (\(t_{\max} - t_{1/2}\)). The required ratio can then be calculated as \((t_{\max} - t_{1/2})/t_{1/2}\).

Step by step solution

01

Understand the scenario

Assume that the maximum height the athlete jumps to be \(y_{\max}\). We know that the time spent in air (hang time) equals to the time taken to go up and down from the maximum height. The task is to find the ratio of time he is above \(y_{\max}/2\) to the time it takes him to go from the floor to \(y_{\max}/2\). This involves considering his journey in two parts: from the floor to \(y_{\max}/2\), then from \(y_{\max}/2\) to the maximum height and back to \(y_{\max}/2\).
02

Calculate the time taken to reach \(y_{\max}/2\) from the floor

Using the kinematic equation \[y_{\max}/2 = v_i*t_{1/2} -1/2* g* t_{1/2}^2\] where \(v_i\) is the initial upward velocity, \(t_{1/2}\) is the time it takes to reach \(y_{\max}/2\) from the floor, and \(g\) is the acceleration due to gravity. Considering the upward motion at maximum height, the final upward velocity is 0. So another kinematic equation can be used, \[0 = v_i - g*t_{\max}\] where \(t_{\max}\) is the time to reach the maximum height. As the hang time is more than 1 second, \(t_{\max} = 1/2\), because the athlete spend half of their time to reach the maximum height and half to descend. Solving these equations will give values of \(v_i\) and \(t_{1/2}\).
03

Calculate the time spent above \(y_{\max}/2\)

The time spent above \(y_{\max}/2\) includes the time taken from \(y_{\max}/2\) to \(y_{\max}\) and the time taken from \(y_{\max}\) to \(y_{\max}/2\). As the ascent and descent are symmetrical in this case, these two times are the same. Therefore, the time above \(y_{\max}/2\) is simply \(t_{\max} - t_{1/2}\).
04

Calculate the required ratio

The required ratio is the time spent above \(y_{\max}/2\) to the time it takes to go from the floor to \(y_{\max}/2\). We have already calculated these two times in step 2 and step 3. So we can calculate the required ratio by dividing the two times.

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Key Concepts

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

Projectile Motion
Projectile motion is a type of motion experienced by an object thrown into the air, under the influence of gravity alone. In the context of a vertical jump, the athlete's movement upward and downward resembles this type of motion. The athlete can be modeled as a particle, moving in a vertical line, to simplify calculations.

When an object is projected, it follows a parabolic path. However, in a vertical jump, we consider only the vertical component of this motion. The initial velocity provided by the athlete during the jump gives the object an upward trajectory, which is eventually curved back down to the ground by gravitational force. The highest point in this trajectory is known as the apex, where the vertical velocity is zero.

Kinematic equations help us predict different aspects of projectile motion such as time of flight, maximum height reached, and the range of the projectile. In the vertical jump problem, equations are used to determine how long the athlete spends in portions of their jump, breaking down the time spent moving upwards and downwards.
Vertical Jump Physics
Physics helps us understand how an athlete's body behaves during a vertical jump. The vertical jump begins from a crouch position, as this helps the athlete generate more power by utilizing their leg muscles. This starting position allows conversion of muscular energy into kinetic energy, propelling the athlete upward.

To simplify our understanding, we treat the athlete as a particle motion problem. Here, the athlete鈥檚 center of mass follows a predictable path, influenced primarily by the initial velocity (generated by the leg muscles) and the gravitational force pulling them back to Earth.

The Role of Initial Velocity

- Initial upward velocity is crucial in determining the maximum height (\(y_{\max}\)) achieved by the jump.- The time to reach this maximum height depends on this initial speed and the acceleration due to gravity.

Time in the Air

- "Hang time," or the time the athlete is airborne, involves both ascent and descent.- Typically, the time to ascend equals the time to descend, making their hang time symmetrical.
Air Resistance
Air resistance, a form of drag force, acts opposite to the direction of motion and is usually considered when calculating real-life projectile motions. However, in many introductory physics problems like the vertical jump scenario, air resistance is often ignored to simplify calculations.

By ignoring air resistance, we assume that only gravity affects the athlete's jump. If air resistance were considered: - The athlete's ascent could slow more quickly than expected, reducing maximum height. - Their descent might be slower, altering the predicted "hang time."
Even though this simplification means solutions won't account for slight deviations due to drag, it helps focus on core principles like initial velocity, maximum height, and symmetrical trajectory.

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Most popular questions from this chapter

You are climbing in the High Sierra when you suddenly find yourself at the edge of a fog-shrouded cliff. To find the height of this cliff, you drop a rock from the top; 8.00 s later you hear the sound of the rock hitting the ground at the foot of the cliff. (a) If you ignore air resistance, how high is the cliff if the speed of sound is \(330 \mathrm{~m} / \mathrm{s} ?\) (b) Suppose you had ignored the time it takes the sound to reach you. In that case, would you have overestimated or underestimated the height of the cliff? Explain.

If the contraction of the left ventricle lasts \(250 \mathrm{~ms}\) and the speed of blood flow in the aorta (the large artery leaving the heart) is \(0.80 \mathrm{~m} / \mathrm{s}\) at the end of the contraction, what is the average acceleration of a red blood cell as it leaves the heart? (a) \(310 \mathrm{~m} / \mathrm{s}^{2} ;(\mathrm{b}) 31 \mathrm{~m} / \mathrm{s}^{2} ;(\mathrm{c}) 3.2 \mathrm{~m} / \mathrm{s}^{2} ;(\mathrm{d}) 0.32 \mathrm{~m} / \mathrm{s}^{2} .\)

A small block has constant acceleration as it slides down a frictionless incline. The block is released from rest at the top of the incline, and its speed after it has traveled \(6.80 \mathrm{~m}\) to the bottom of the incline is \(3.80 \mathrm{~m} / \mathrm{s}\). What is the speed of the block when it is \(3.40 \mathrm{~m}\) from the top of the incline?

Two stones are thrown vertically upward from the ground, one with three times the initial speed of the other. (a) If the faster stone takes \(10 \mathrm{~s}\) to return to the ground, how long will it take the slower stone to return? (b) If the slower stone reaches a maximum height of \(H,\) how high (in terms of \(H\) ) will the faster stone go? Assume free fall.

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof \(1.00 \mathrm{~s}\) later. Ignore air resistance. (a) If the height of the building is \(20.0 \mathrm{~m},\) what must the initial speed of the first ball be if both are to hit the ground at the same time? On the same graph, sketch the positions of both balls as a function of time, measured from when the first ball is thrown. Consider the same situation, but now let the initial speed \(v_{0}\) of the first ball be given and treat the height \(h\) of the building as an unknown. (b) What must the height of the building be for both balls to reach the ground at the same time if (i) \(v_{0}\) is \(6.0 \mathrm{~m} / \mathrm{s}\) and (ii) \(v_{0}\) is \(9.5 \mathrm{~m} / \mathrm{s} ?\) (c) If \(v_{0}\) is greater than some value \(v_{\max },\) no value of \(h\) exists that allows both balls to hit the ground at the same time. Solve for \(v_{\max }\). The value \(v_{\max }\) has a simple physical interpretation. What is it? (d) If \(u_{0}\) is less than some value \(v_{\min }\), no value of \(h\) exists that allows both balls to hit the ground at the same time. Solve for \(v_{\min }\). The value \(v_{\min }\) also has a simple physical interpretation. What is it?
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