/*! 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 69 A student standing on the ground... [FREE SOLUTION] | 91Ó°ÊÓ

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A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of \(15 \mathrm{m} / \mathrm{s}\) when the hand is \(2.0 \mathrm{m}\) above the ground. How long is the ball in the air before it hits the ground? (The student moves her hand out of the way.)

Short Answer

Expert verified
The ball stays in the air for approximately 2.17 seconds before it hits the ground.

Step by step solution

01

Compute Time for Upward Journey

The time taken for the upward journey can be calculated using the formula \( t = \frac{v}{g} \), where \( v \) is the speed of the ball and \( g \) is acceleration due to gravity. In this case, \( v = 15 \, \mathrm{m/s} \) and \( g = 9.8 \, \mathrm{m/s}^2 \). Substituting these into the formula, we have \( t = \frac{15}{9.8} = 1.53 \, s \).
02

Compute Time for Downward Journey

The time taken for the journey downwards can be calculated using the formula \( t = \sqrt{2h/g} \), where \( h \) is the height from which the ball is thrown. The provided height is \( 2.0 \, m \). Substituting this value and the acceleration due to gravity into the formula, we have \( t = \sqrt{2*2/9.8} = 0.64 \, s \).
03

Compute Total Time

The total time the ball is in the air is the sum of the time for the upward journey and the time for the downward journey. Therefore, the total time is \( 1.53 \, s + 0.64 \, s = 2.17 \, s \).

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Key Concepts

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

Kinematics
Kinematics is a branch of physics that describes the motion of objects without considering the forces that cause the motion. It focuses on understanding concepts such as velocity, acceleration, and displacement. In this exercise, the ball thrown upwards falls under the category of kinematics as we are analyzing its movement.

One kinematic equation often used is:
  • \( v = u + at \)
  • \( s = ut + \frac{1}{2}at^2 \)
  • \( v^2 = u^2 + 2as \)
Where:
  • \( v \) is the final velocity
  • \( u \) is the initial velocity
  • \( a \) is acceleration
  • \( s \) is displacement
  • \( t \) is time
By understanding these equations, you can solve diverse motion problems, like determining how long the ball is in the air.
The straightforward calculations done step by step in the example indicate how each part of the motion is considered separately, calculating the time taken for upward and downward journeys independently.
Gravity
Gravity is the force that attracts two masses towards each other. On Earth, it gives weight to physical objects and causes them to fall towards the ground when dropped. It is crucial in projectile motion problems because it influences the path and time of flight of any projectile.

The gravitational constant \( g \) on Earth is approximately \( 9.8 \, \mathrm{m/s}^2 \), and it acts downwards towards the center of the planet. In this exercise, gravity is responsible for slowing the ball as it travels upwards and accelerating it downwards after reaching its peak.
The equation used to determine the upward journey time \( t = \frac{v}{g} \) showcases how gravity influences the timeline of a projectile's motion. The constant downward pull ensures the projectile eventually returns to the ground while also determining the highest point reached in its trajectory. Understanding gravity's role helps you predict the motion of an object thrown or projected upwards.
Vertical Motion
Vertical motion refers to the motion of an object moving up and down along the vertical axis. This kind of motion is significantly influenced by gravity, especially in the absence of air resistance. In the exercise, the ball is projected vertically upwards, which is a classic example of vertical motion.

To understand vertical motion better:
  • The initial velocity \( v_i \) determines how high the projectile travels.
  • Gravity decelerates the projectile until it stops at the peak (when velocity = 0) before accelerating it downwards.
  • The time taken to reach the maximum height mirroring closely the time to descend to the original height "plus" any initial height, such as the student's hand at \( 2.0 \mathrm{m} \).
This uniform motion makes it easier to calculate various factors like time in the air and the maximum height. In problems of this nature, it's important to separately calculate each segment of motion—the ascent and descent—because they provide crucial data for the total time an object remains airborne. The integration of the upward and downward journey times provides a complete picture of the projectile's flight duration.

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Most popular questions from this chapter

A Porsche challenges a Honda to a \(400 \mathrm{m}\) race. Because the Porsche's acceleration of \(3.5 \mathrm{m} / \mathrm{s}^{2}\) is larger than the Honda's \(3.0 \mathrm{m} / \mathrm{s}^{2},\) the Honda gets a \(100-\mathrm{m}\) head start-it is only \(300 \mathrm{m}\) from the finish line. Assume, somewhat unrealistically, that both cars can maintain these accelerations the entire distance. Who wins, and by how much time?

If Chameleons can rapidly project their very long tongues to catch nearby insects. The tongue of the tiny Rosette-nosed chameleon has the highest acceleration of a body part of any amniote (reptile, bird, or mammal) ever measured. In a somewhat simplified model of its tongue motion, the tongue, starting from rest, first undergoes a constant-acceleration phase with an astounding magnitude of \(2500 \mathrm{m} / \mathrm{s}^{2} .\) This acceleration brings the tongue up to a final speed of \(5.0 \mathrm{m} / \mathrm{s}\). It continues at this speed for \(22 \mathrm{ms}\) until it hits its target. a. How long does the acceleration phase last? b. What is the total distance traveled by the chameleon's tongue?

In an action movie, the villain is rescued from the ocean by grabbing onto the ladder hanging from a helicopter. He is so intent on gripping the ladder that he lets go of his briefcase of counterfeit money when he is \(130 \mathrm{m}\) above the water. If the briefcase hits the water \(6.0 \mathrm{s}\) later, what was the speed at which the helicopter was ascending?

Alan leaves Los Angeles at 8: 00 AM to drive to San Francisco, \(400 \mathrm{mi}\) away. He travels at a steady \(50 \mathrm{mph}\). Beth leaves Los Angeles at 9: 00 AM and drives a steady 60 mph. a. Who gets to San Francisco first? b. How long does the first to arrive have to wait for the second?

a. What constant acceleration, in SI units, must a car have to go from zero to 60 mph in 10 s? b. What fraction of \(g\) is this? c. How far has the car traveled when it reaches \(60 \mathrm{mph} ?\) Give your answer both in \(\mathrm{SI}\) units and in feet.

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