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Chapter 2: Problem 38

You throw a glob of putty straight up toward the ceiling, which is 3.60 m above the point where the putty leaves your hand. The initial speed of the putty as it leaves your hand is 9.50 m/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
(a) 4.43 m/s; (b) 0.517 s.

Step by step solution

01

Determine Final Velocity

We can use the kinematic equation \( v^2 = u^2 + 2as \) to find the final velocity of the putty just before it strikes the ceiling. Here, \( v \) is the final velocity, \( u = 9.50 \, \text{m/s} \) is the initial velocity, \( a = -9.81 \, \text{m/s}^2 \) is the acceleration due to gravity (negative because it's acting downward), and \( s = 3.60 \, \text{m} \) is the displacement. Substitute the values into the equation:\[ v^2 = (9.50)^2 + 2(-9.81)(3.60) \]\[ v^2 = 90.25 - 70.632 \]\[ v^2 = 19.618 \]\[ v = \sqrt{19.618} = 4.43 \, \text{m/s} \]Hence, the speed of the putty just before it strikes the ceiling is 4.43 m/s.
02

Calculate Time to Reach Ceiling

To find the time taken for the putty to reach the ceiling, we use the kinematic equation \( v = u + at \), where \( v \) is the final velocity from the previous step, \( u = 9.50 \, \text{m/s} \), \( a = -9.81 \, \text{m/s}^2 \), and \( t \) is the time. Rearrange to solve for \( t \):\[ t = \frac{v - u}{a} \]Substitute the values:\[ t = \frac{4.43 - 9.50}{-9.81} \]\[ t = \frac{-5.07}{-9.81} \]\[ t = 0.517 \, \text{s} \]Therefore, the putty takes approximately 0.517 seconds to reach the ceiling.

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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 the motion experienced by an object thrown into the air, subject to only the acceleration due to gravity. When a projectile is launched, it follows a curved path under the influence of gravity alone. In our scenario, the glob of putty is thrown upwards, making its motion an example of one-dimensional projectile motion.

There are a few characteristics of projectile motion:
  • The motion has both vertical and horizontal components, but in this example, we only consider the vertical component since the putty moves straight up.
  • The vertical motion behaves uniformly due to gravity acting as a constant downward force.
  • At the peak of its motion (just before hitting the ceiling), the vertical velocity momentarily becomes zero before gravity pulls the object down again.
Understanding these components helps in analyzing how the putty behaves from the moment it leaves your hand until it strikes the ceiling.
Acceleration Due to Gravity
The acceleration due to gravity is a key factor in any projectile motion scenario. It acts as a constant force pulling the object back towards the Earth. This acceleration is symbolized as "g," and its standard value at the Earth's surface is approximately \(-9.81 \, \text{m/s}^2\) (the negative sign indicates a downward force).

In the exercise at hand:
  • "g" slows down the upward ascent of the putty until it halts its rise.
  • Subsequently, it accelerates the putty towards the ceiling, contributing to its final velocity just before impact.
  • The constant value of \(9.81 \, \text{m/s}^2\) allows us to use standard kinematic equations effectively in our calculations.
Proper comprehension of gravity's impact on a projectile is vital. It helps in predicting the behavior of any object in free fall or thrown vertically.
Kinematic Equations
Kinematic equations are essential tools for solving problems involving uniform acceleration vector motion, such as projectile motion. These equations connect different physical quantities like velocity, displacement, time, and acceleration.

Key kinematic equations include:
  • \(v = u + at\)- Relates final velocity \(v\), initial velocity \(u\), acceleration \(a\), and time \(t\).
  • \(v^2 = u^2 + 2as\)- Useful for finding the final velocity without knowing the time directly.
  • \(s = ut + \frac{1}{2}at^2\)- Calculates displacement \(s\), which is integrated over time from initial conditions.
In the problem, the equation \(v^2 = u^2 + 2as\) helped determine the speed of the putty just before it hit the ceiling. The time was then calculated using \(v = u + at\). By mastering these equations, one can solve a broad range of kinematics-related problems effortlessly.

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Most popular questions from this chapter

A hot-air balloonist, rising vertically with a constant velocity of magnitude 5.00 m/s, releases a sandbag at an instant when the balloon is 40.0 m above the ground (\(\textbf{Fig. E2.44}\)). After the sandbag is released, it is in free fall. (a) Compute the position and velocity of the sandbag at 0.250 s and 1.00 s after its release. (b) How many seconds after its release does the bag strike the ground? (c) With what magnitude of velocity does it strike the ground? (d) What is the greatest height above the ground that the sandbag reaches? (e) Sketch \(a_y-t\), \(v_y-t\), and \(y-t\) graphs for the motion.

A rocket starts from rest and moves upward from the surface of the earth. For the first 10.0 s of its motion, the vertical acceleration of the rocket is given by \(a_y =\) (2.80 m/s\(^3)t\), where the \(+y\)-direction is upward. (a) What is the height of the rocket above the surface of the earth at \(t =\) 10.0 s? (b) What is the speed of the rocket when it is 325 m above the surface of the earth?

At launch a rocket ship weighs 4.5 million pounds. When it is launched from rest, it takes 8.00 s to reach 161 km/h; at the end of the first 1.00 min, its speed is 1610 km/h. (a) What is the average acceleration (in m/s\(^2\)) of the rocket (i) during the first 8.00 s and (ii) between 8.00 s and the end of the first 1.00 min? (b) Assuming the acceleration is constant during each time interval (but not necessarily the same in both intervals), what distance does the rocket travel (i) during the first 8.00 s and (ii) during the interval from 8.00 s to 1.00 min?

The rocket-driven sled \(\textit{Sonic Wind No. 2,}\) used for investigating the physiological effects of large accelerations, runs on a straight, level track 1070 m (3500 ft) long. Starting from rest, it can reach a speed of 224 m/s(500 mi/h) in 0.900 s. (a) Compute the acceleration in m/s\(^2\), assuming that it is constant. (b) What is the ratio of this acceleration to that of a freely falling body (\(g\))? (c) What distance is covered in 0.900 s? (d) A magazine article states that at the end of a certain run, the speed of the sled decreased from 283 m/s (632 mi/h) to zero in 1.40 s and that during this time the magnitude of the acceleration was greater than 40\(g\). Are these figures consistent?

A car travels in the \(+x\)-direction on a straight and level road. For the first 4.00 s of its motion, the average velocity of the car is \(v_{av-x}\) = 6.25 m/s. How far does the car travel in 4.00 s?
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