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Chapter 2: Problem 34

You throw a glob of putty straight up toward the ceiling. which is \(3.60 \mathrm{~m}\) above the point where the putty leaves your hand. The initial speed of the putty as it leaves your hand is \(9.50 \mathrm{~m} / \mathrm{s}\). (a) What is the speed of the putty just before it strikes the ceiling? (b) How much time from when it leaves your hand does it take the putty to reach the ceiling?

Short Answer

Expert verified
The speed of the putty just before it strikes the ceiling is 8.60 m/s, and it takes 0.73 seconds for the putty to reach the ceiling.

Step by step solution

01

Identify Given Inputs

From the problem the following values are given: initial velocity (u) as 9.50 m/s, final distance (s) as 3.60 m, and gravitational acceleration (g) as 9.8 m/s虏 (value is known, not given in the problem). Note the acceleration due to gravity is acting downward, therefore it's taken negative.
02

Solve for Final Velocity

To solve for final velocity, we use the kinematic equation, \(v^{2} = u^{2} + 2gs\). Substituting the values we have: \(v^{2} = (9.50\,m/s)^{2} + 2*(-9.8\,m/s虏)*(3.60\, m)\). Simplifying this equation will give the square of velocity. Taking the square root of that quantity gives the magnitude of the velocity. This is the velocity of the putty when it just before hits the ceiling.
03

Solve for Time

To find the time it takes the putty to reach the ceiling, we use the second equation of motion: \(s = ut + 0.5gt^{2}\). Rearranging for \(t\), and substituting the given values, we get: \(t = \sqrt{2s/g}\). Solve for the time.

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Key Concepts

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

Initial Velocity
When discussing the motion of objects, especially in kinematics, the initial velocity is a vital concept. The initial velocity (usually denoted as \( u \)) represents the speed at which an object begins its motion.
In the case of the putty exercise, the initial velocity is given as 9.50 m/s. This means when the putty leaves the hand, it starts its journey toward the ceiling at a speed of 9.50 meters per second.
This value is crucial because it serves as the starting point for all subsequent calculations involving the object's motion. Knowing the initial velocity allows us to use kinematic equations to predict other properties of the motion, like final velocity and time taken to reach a certain point. By understanding and correctly identifying the initial velocity, we can accurately analyze the trajectory of the object.
Gravitational Acceleration
Gravitational acceleration, often denoted as \( g \), is the acceleration experienced by an object due to the force of gravity. For objects close to the Earth's surface, this acceleration is approximately 9.8 m/s虏.
In this exercise, gravity acts upon the putty when it is thrown upwards toward the ceiling. Importantly, the direction of gravity is downward, opposite to the throw. Therefore, in calculations, gravitational acceleration is considered as a negative value (\( -9.8 \) m/s虏) to reflect this direction.
  • It affects how fast the object slows down while traveling upward.
  • It determines how quickly the object will accelerate back downwards if thrown upward.
Understanding gravitational acceleration is essential to predicting the behavior of any object in freefall or projectile motion. It's a constant parameter in all kinematic equations related to vertical motion near Earth's surface.
Kinematic Equations
Kinematic equations are the backbone of solving motion problems in physics. These equations relate the different variables involved in kinematic situations, such as initial velocity, final velocity, acceleration, displacement, and time.
For the putty example, two key kinematic equations are used:
  • Final velocity (\( v \)): \( v^2 = u^2 + 2gs \)
  • Time of flight (\( t \)): \( s = ut + 0.5gt^2 \)
Both equations allow us to solve for unknown variables. In this problem, we used them to find the putty's speed right before hitting the ceiling and the time it takes to reach the ceiling.
By plugging in known values and solving these equations, we can predict how the putty will behave as it travels. These equations are important tools for anyone learning physics, as they unravel the mystery of how objects move under the influence of forces and accelerations.

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Most popular questions from this chapter

A student is running at her top speed of \(5.0 \mathrm{~m} / \mathrm{s}\) to catch a bus, which is stopped at the bus stop. When the student is still \(40.0 \mathrm{~m}\) from the bus, it starts to pull away, moving with a constant acceleration of \(0.170 \mathrm{~m} / \mathrm{s}^{2}\). (a) For how much time and what distance does the student have to run at \(5.0 \mathrm{~m} / \mathrm{s}\) before she overtakes the bus? (b) When she reaches the bus, how fast is the bus traveling? (c) Sketch an \(x-t\) graph for both the student and the bus. Take \(x=0\) at the initial position of the student. (d) The equations you used in part (a) to find the time have a second solution, corresponding to a later time for which the student and bus are again at the same place if they continue their specified motions. Explain the significance of this second solution. How fast is the bus traveling at this point? (e) If the student's top speed is \(3.5 \mathrm{~m} / \mathrm{s},\) will she catch the bus? (f) What is the minimum speed the student must have to just catch up with the bus? For what time and what distance does she have to run in that case?

During your summer internship for an aerospace company, you are asked to design a small research rocket. The rocket is to be launched from rest from the earth's surface and is to reach a maximum height of \(960 \mathrm{~m}\) above the earth's surface. The rocket's engines give the rocket an upward acceleration of \(16.0 \mathrm{~m} / \mathrm{s}^{2}\) during the time \(T\) that they fire. After the engines shut off, the rocket is in free fall. Ignore air resistance. What must be the value of \(T\) in order for the rocket to reach the required altitude?

You throw a rock straight up and find that it returns to your hand \(3.60 \mathrm{~s}\) after it left your hand. Neglect air resistance. What was th maximum height above your hand that the rock reached?

A race car starts from rest and travels east along a straight and level track. For the first \(5.0 \mathrm{~s}\) of the car's motion, the castward component of the car's velocity is given by \(v_{x}(t)=\left(0.860 \mathrm{~m} / \mathrm{s}^{3}\right) t^{2}\) What is the acceleration of the car when \(v_{x}=12.0 \mathrm{~m} / \mathrm{s} ?\)

A turtle crawls along a straight line, which we'll call the \(x\) -axis with the positive direction to the right. The equation for the turtle's position as a function of time is \(x(t)=50.0 \mathrm{~cm}+\) \((2.00 \mathrm{~cm} / \mathrm{s}) t-\left(0.0625 \mathrm{~cm} / \mathrm{s}^{2}\right) t^{2} .\) (a) Find the turtle's initial velocity. initial position, and initial acceleration. (b) At what time \(t\) is the velocity of the turtle zero? (c) How long after starting does it take the turtle to return to its starting point? (d) At what times \(t\) is the turtle a distance of \(10.0 \mathrm{~cm}\) from its starting point? What is the velocity (magnitude and direction) of the turtle at each of those times? (e) Sketch graphs of \(x\) versus \(t, v_{1}\) versus \(t,\) and \(a_{f}\) versus \(t,\) for the time interval \(t=0\) to \(t=40 \mathrm{~s}\)
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