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Chapter 4: Problem 39

A Standing Vertical Jump. Basketball player Darrell Griffith is on record as attaining a standing vertical jump of 1.2 \(\mathrm{m}(4 \mathrm{ft}) .\) (This means that he moved upward by 1.2 \(\mathrm{m}\) after his feet left the floor.) Griffith weighed 890 \(\mathrm{N}(200 \mathrm{lb}) .\) (a) What is his speed as he leaves the floor? (b) If the time of the part of the jump before his feet left the floor was 0.300 s, what was his average acceleration (magnitude and direction) while he was pushing against the floor? (c) Draw his free-body diagram (see Section 4.6\()\) . In terms of the forces on the diagram, what is the net force on him? Use Newton's laws and the results of part (b) to calculate the aver- age force he applied to the ground.

Short Answer

Expert verified
(a) 4.85 m/s, (b) 16.17 m/s虏 upwards, (c) 2358 N upwards.

Step by step solution

01

Determine Initial Speed Using Kinematics

To find the initial speed \( v_0 \) as Darrell leaves the floor, use the kinematic equation: \( v^2 = v_0^2 + 2a(y - y_0) \). Since the final speed at the top of the jump is 0 (\( v = 0 \)), and \( a = -9.8 \, \text{m/s}^2 \) (due to gravity), and the displacement (\( y - y_0 = 1.2 \, \text{m} \)), we get:\0 = v_0^2 - 2 \times 9.8 \times 1.2.\Solve for \( v_0 \):\v_0^2 = 2 \times 9.8 \times 1.2 = 23.52,\v_0 = \sqrt{23.52} \approx 4.85 \, \text{m/s}.
02

Calculate Average Acceleration

Using the initial speed from Step 1 and the time given (0.300 s), apply the formula \( v = v_0 + at \) to find the average acceleration. Here, the velocity just before leaving the ground is \( v_0 \), and initial velocity is 0. Rearrange to find \( a \):\4.85 = 0 + a \times 0.300,\a = \frac{4.85}{0.300} \approx 16.17 \, \text{m/s}^2.\This is the magnitude of the average acceleration, and the direction is upwards.
03

Analyze Forces and Find Net Force

In the free-body diagram (FBD), two forces act on Darrell: the gravitational force (\( F_g = 890 \, \text{N} \)) downward and the normal force (\( F_N \)) from the ground upward. The net force \( F_{net} \) is \( m \cdot a \), where \( m \) is his mass. Using Newton鈥檚 second law, \( m = \frac{F_g}{g} = \frac{890}{9.8} \approx 90.8 \, \text{kg} \).\So, \( F_{net} = 90.8 \times 16.17 = 1467.73 \, \text{N} \).
04

Calculate Average Force Applied to the Ground

Since \( F_{net} = F_N - F_g \), solve for \( F_N \):\\( F_N = F_{net} + F_g = 1467.73 + 890 = 2357.73 \, \text{N} \).\Thus, the average force applied to the ground by Darrell is approximately 2358 N upward.

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Key Concepts

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

vertical jump
A vertical jump is a movement that requires an individual to leap straight up into the air. This is often measured from the standing ground level to the highest point reached. The height of a vertical jump, such as Darrell Griffith's impressive 1.2-meter jump, can tell us a lot about the jumper's initial speed and the forces at play. When analyzing a vertical jump, several physics concepts come into play, including kinematics, force, and acceleration. The jump's efficacy can be determined by calculating the jumper's speed as they leave the floor and the force they apply. Typically, a higher vertical jump indicates greater initial speed and force.
free-body diagram
A Free-Body Diagram (FBD) is a simple way to visualize the forces acting on an object or person. In Darrell Griffith's jump, the FBD helps us understand the forces at play during the jump. The diagram would show two main forces: the downward gravitational force (\(F_g = 890 \, \text{N}\)) and the upward normal force (\(F_N\)) exerted by the ground. These forces are crucial in determining the net force on Darrell as he pushes off the ground. Understanding these forces allows us to better analyze the dynamics of motion during the jump.
Newton's laws
Newton's laws of motion are foundational for understanding dynamics, and they are quite applicable in analyzing Darrell Griffith's vertical jump. The second law, in particular, states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration (\(F_{net} = m \cdot a\)). This law allows us to calculate the net force when the mass and acceleration are known. In Griffith's jump, this principle is key as it helps us determine how his acceleration (upwards) and the force from the ground combined to propel him into the air.
acceleration calculation
Acceleration is a measure of how quickly velocity changes and is crucial in the context of a jump. To find Darrell Griffith's acceleration during the jump, we utilize the kinematic equations. Initially, we calculate his speed upon leaving the ground using the formula: \(v^2 = v_0^2 + 2a(y - y_0)\). By knowing the jump height and that the final speed is zero at the highest point, we can rearrange and solve for his initial speed as 4.85 m/s.
Once the speed is known, we use the formula \(v = v_0 + at\) to calculate the average acceleration while his feet push off the floor. With the given time of 0.300 s, we find that the average upward acceleration is approximately 16.17 m/s\(^2\). This value helps evaluate how effectively the jumper transmuted muscular force into vertical acceleration.

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Most popular questions from this chapter

Two forces, \(\vec{\boldsymbol{F}}_{1}\) and \(\vec{\boldsymbol{F}}_{2},\) a point. The magnitude of \(\vec{\boldsymbol{F}}_{1}\) is \(9.00 \mathrm{N},\) and its direction is \(60.0^{\circ}\) above the \(x\) -axis in the second quadrant. The magnitude of \(\vec{\boldsymbol{F}}_{2}\) is \(6.00 \mathrm{N},\) and its direction is \(53.1^{\circ}\) below the \(x\) -axis in the third quadrant. (a) What are the \(x\) - and \(y\) -components of the resultant force? (b) What is the magnitude of the resultant force?

A skier of mass 65.0 \(\mathrm{kg}\) is pulled up a snow-covered slope at constant speed by a tow rope that is parallel to the ground. The ground slopes upward at a constant angle of \(26.0^{\circ}\) above the horizontal, and you can ignore friction. (a) Draw a clearly labeled free body diagram for the skier. (b) Calculate the tension in the tow rope.?

A loaded elevator with very worn cables has a total mass of \(2200 \mathrm{kg},\) and the cables can withstand a maximum tension of \(28,000 \mathrm{N},(\mathrm{a})\) Draw the free-body force diagram for the elevator. In terms of the forces on your diagram, what is the net force on the elevator? Apply Newton's second law to the elevator and find the maximum upward acceleration for the elevator if the cables are not to break. (b) What would be the answer to part (a) if the elevator were on the moon, where \(g=1.62 \mathrm{m} / \mathrm{s}^{2} ?\)

An oil tanker's engines have broken down, and the wind is blowing the tanker straight toward a reef at a constant speed of 1.5 \(\mathrm{m} / \mathrm{s}(\) Fig. \(\mathrm{P} 4.38) .\) When the tanker is 500 \(\mathrm{m}\) from the reef, the wind dies down just as the engineer gets the engines going again. The rudder is stuck, so the only choice is to try to accelerate straight backward away from the reef. The mass of the tanker and cargo is \(3.6 \times 10^{7} \mathrm{kg}\) , and the engines produce a net horizontal force of \(8.0 \times 10^{4} \mathrm{N}\) on the tanker. Will the ship hit the reef? If it does, will the oil be sate? The hull can withstand an impact at a speed of 0.2 \(\mathrm{m} / \mathrm{s}\) or less. You can ignore the retarding force of the water on the tanker's hull.

A box rests on a frozen pond, which serves as a frictionless horizontal surface. If a fisherman applies a horizontal force with magnitude 48.0 \(\mathrm{N}\) to the box and produces an acceleration of mag-nitude \(3,00 \mathrm{m} / \mathrm{s}^{2},\) what is the mass of the box?
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