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Chapter 4: Problem 51

A standing vertical jump. Basketball player Darrell Griffith is on record as attaining a standing vertical jump of 1.2 \(\mathrm{m}\) (4 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 was his speed as he left the floor? (b) If the time of the part of the jump before his feet left the floor was 0.300 s, what were the magnitude and direction of his acceleration (assuming it to be constant) while he was pushing against the floor? (c) Draw a free-body diagram of Griffith during the jump. (d) Use Newton's laws and the results of part (b) to calculate the force he applied to the ground during his jump.

Short Answer

Expert verified
(a) 4.85 m/s; (b) 16.17 m/s虏 upwards; (c) Forces: weight downward, normal force upward; (d) Applied force: 2357 N.

Step by step solution

01

Finding Final Velocity

To find Darrell Griffith's speed as he left the floor, we use the kinematic equation: \[ v^2 = u^2 + 2as \]Where \( v \) is the final velocity, \( u \) is the initial velocity (0 m/s since he starts from rest), \( a \) is the acceleration (which is gravity, \(-9.8\, \text{m/s}^2\) when moving upward), and \( s \) is the displacement (1.2 m). Solving for \( v \):\[ v^2 = 0 + 2 \times 9.8 \times 1.2 \]\[ v^2 = 23.52 \]\[ v = \sqrt{23.52} \approx 4.85 \, \text{m/s} \]Thus, his speed as he left the floor was approximately 4.85 m/s.
02

Calculating Acceleration

Using the velocity obtained from step 1 and the time given (0.300 s), we calculate the acceleration using the formula: \[ v = u + at \]where \( u = 0 \), \( t = 0.3 \, \text{s} \), and \( v = 4.85 \, \text{m/s} \). Solving for \( a \):\[ 4.85 = 0 + a \times 0.3 \]\[ a = \frac{4.85}{0.3} \approx 16.17 \, \text{m/s}^2 \]The magnitude of his acceleration while pushing against the floor was approximately 16.17 m/s虏 upward.
03

Free-body Diagram

In the free-body diagram for Darrell during the jump, two main forces act on him: his weight \( W = 890 \, \text{N} \) downwards and the normal force \( F_N \) exerted by the ground upwards. During the push-off phase, the normal force is larger than his weight, enabling the upward acceleration.
04

Calculating Applied Force

Apply Newton's second law to find the force Darrell applied to the ground. The net force \( F_{net} \) during the acceleration phase is:\[ F_{net} = ma \]where \( m = \frac{890}{9.8} = 90.82 \, \text{kg} \), and \( a = 16.17 \, \text{m/s}^2 \). So, \( F_{net} = 90.82 \times 16.17 \approx 1467 \, \text{N} \).The normal force is the athlete's weight plus the net force (since both act upward against the gravitational force): \[ F_N = 1467 + 890 = 2357 \, \text{N} \]Thus, Darrell applied approximately 2357 N of force to the ground.

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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 in understanding motion, especially when analyzing vertical jumps in physics. These equations relate the motion parameters: initial velocity, final velocity, acceleration, distance, and time. In Darrell Griffith's jump, the kinematic equation \[ v^2 = u^2 + 2as \]was employed to find the speed at which he left the floor. Here, the initial velocity (\( u \)) was 0 m/s since he started from rest. The acceleration (\( a \)) is due to gravity, which acts downwards with \( -9.8 \, \text{m/s}^2 \), while the displacement (\( s \)) was his jump height of 1.2 m. Solving for the final velocity (\( v \)) provided the speed needed to achieve the given vertical height. This application of kinematic equations shows how they help break down complex motion into understandable components.
Newton's Laws
Newton's Laws, particularly the second law, are central to calculating forces involved during a vertical jump. The second law states that the net force acting on an object is equal to the product of its mass (\( m \)) and its acceleration (\( a \)). Symbolically, this is expressed as:\[ F_{net} = ma \]For Darrell's jump, his mass was derived from his weight (\( W = 890 \, \text{N} \)), leading to a mass of about 90.82 kg since \( mass = \frac{weight}{gravity} \). Applying the found acceleration of 16.17 \( \text{m/s}^2 \), we calculated the net force he exerted. Newton's laws help connect the dots between motion and the forces causing it, allowing us to determine how much force Darrell needed to push off the ground.
Free-Body Diagram
A free-body diagram is a simple drawing that depicts all forces acting upon an object, essential for visualizing and solving physics problems. In the case of the vertical jump, Darrell Griffith is subjected to two main forces during his take-off:
  • His weight (\( W = 890 \, \text{N} \)): a downward force due to gravity.
  • The normal force (\( F_N \)): an upward force exerted by the ground.
The free-body diagram is insightful as it shows the interaction of these forces. When Darrell begins his jump, the normal force must exceed his weight to provide the required upward acceleration. This upward force propels him off the ground, overcoming gravity. By analyzing the forces present in a free-body diagram, one can determine the behavior of objects in motion under various force influences.
Acceleration Calculation
Calculating acceleration involves determining how quickly velocity changes over time. During a vertical jump, knowing the acceleration helps us understand the dynamics of take-off. Darrell's acceleration while pushing against the floor was calculated using\[ v = u + at \]With an initial velocity (\( u = 0 \)), final velocity (\( v = 4.85 \, \text{m/s} \)), and time of take-off (\( t = 0.3 \) seconds), the acceleration (\( a \)) came out to about 16.17 \( \text{m/s}^2 \). This value represents how forceful the push-off was, describing the rapid gain in velocity needed to reach a height of 1.2 meters. Understanding acceleration highlights the effects of forces and the nature of motion in a jump.

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Most popular questions from this chapter

A person throws a 2.5 lb stone into the air with an initial upward speed of 15 \(\mathrm{ft} / \mathrm{s}\) . Make a free-body diagram for this stone (a) after it is free of the person's hand and is traveling upward, (b) at its highest point, (c) when it is traveling downward, and (d) while it is being thrown upward, but is still in contact with the person's hand.

Human biomechanics. The fastest pitched baseball was measured at 46 \(\mathrm{m} / \mathrm{s}\) . Typically, a baseball has a mass of 145 \(\mathrm{g}\) . If the pitcher exerted his force (assumed to be horizontal and constant) over a distance of \(1.0 \mathrm{m},(\mathrm{a})\) what force did he produce on the ball during this record-setting pitch? (b) Make free-body diagrams of the ball during the pitch and just after it has left the pitcher's hand.

An electron \(\left(\mathrm{mass}=9.11 \times 10^{-31} \mathrm{kg}\right)\) leaves one end of a TV picture tube with zero initial speed and travels in a straight line to the accelerating grid, which is 1.80 \(\mathrm{cm}\) away. It reaches the grid with a speed of \(3.00 \times 10^{6} \mathrm{m} / \mathrm{s} .\) If the accelerating force is constant, compute (a) the acceleration of the electron, (b) the time it takes the electron to reach the grid, and (c) the net force that is accelerating the electron, in newtons. (You can ignore the gravitational force on the electron.)

The driver of a 1750 \(\mathrm{kg}\) car traveling on a horizontal road at 110 \(\mathrm{km} / \mathrm{h}\) suddenly applies the brakes. Due to a slippery pavement, the friction of the road on the tires of the car, which is what slown the car, is 25\(\%\) of the weight of the car. (a) What is the acceleration of the car? (b) How many meters does it travel before stopping under these conditions?

What does a 138 \(\mathrm{N}\) rock weigh if it is accelerating (a) upward at \(12 \mathrm{m} / \mathrm{s}^{2},\) (b) downward at 3.5 \(\mathrm{m} / \mathrm{s}^{2} ?\) (c) What would be the answers to parts (a) and (b) if the rock had a mass of 138 \(\mathrm{kg} ?(\mathrm{d})\) What would be the answers to parts (a) and (b) if the rock were moving with a constant upward velocity of 23 \(\mathrm{m} / \mathrm{s} ?\)
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