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Chapter 2: Problem 31

(a) If a flea can jump straight up to a height of \(0.440 \mathrm{~m}\), what is its initial speed as it leaves the ground? (b) How long is it in the air?

Short Answer

Expert verified
The initial speed of the flea as it leaves the ground is around \(2.94 \mathrm{~m/s}\). The flea is in the air for about \(0.60 \mathrm{~s}\).

Step by step solution

01

Compute the initial speed of the flea

First, apply the second equation of motion which states: \[v_f^2 = v_i^2 + 2a d,\]with \(v_f\) as the final velocity, \(v_i\) as the initial velocity, \(a\) as acceleration, and \(d\) as displacement. Here, when the flea hits the peak of the jump, it momentarily stops, so \(v_f = 0\), \(a\) is the acceleration due to gravity which is \(-9.81 \mathrm{~m/s}^2\), \(d\) is the height reached by the flea \(0.440 \mathrm{~m}\). So, \[0 = v_i^2 + 2(-9.81)(0.440)\].Solving for \(v_i\) will give the initial speed.
02

Calculate the time the flea is in the air

Next, apply the first equation of motion which states:\[ v_f = v_i + a t,\]where \(v_f\) is the final velocity, \(v_i\) is the initial velocity, \(a\) is acceleration, and \(t\) is time.When the flea lands after the jump, its final velocity is equal but opposite to the initial velocity, \(v_f = -v_i\). Here, \(a\) is the acceleration due to gravity which is \(-9.81 \mathrm{~m/s}^2\).So,\[ -v_i = v_i + (-9.81)t.\]Solving for \(t\) will give the time the flea is in the air.

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Key Concepts

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

Equations of Motion
Understanding the equations of motion is crucial for analyzing objects' movement in physics. These equations enable us to predict an object's position or state of motion at a given time, provided we have relevant initial conditions and constant acceleration. One of the most commonly used equations of motion is \[ v_f^2 = v_i^2 + 2a d, \] where:\
    \
  • \(v_f\) is the final velocity,\
    • \
  • \(v_i\) is the initial velocity,\
    • \
  • \(a\) is the constant acceleration,\
    • \
  • \(d\) is the displacement.\
    • \
\
In the context of a flea's jump, as it reaches the peak height, the final velocity becomes zero, which simplifies our equation and allows us to solve for the initial speed or velocity.
Initial Velocity
Initial velocity, denoted by \(v_i\), is the speed at which an object starts its journey. When a flea jumps up, it leaves the ground with an initial velocity that can be calculated using the appropriate motion equation. In our problem, we are provided with the displacement (the maximum height reached by the flea), and we know the acceleration due to gravity. By setting the final velocity to zero at the peak height and rearranging the equation of motion, we can solve for the initial velocity, which gives us insight into how fast the flea was moving when it left the ground.
Acceleration Due to Gravity
The acceleration due to gravity, typically represented by \(g\), is a constant value of approximately \(-9.81 \mathrm{~m/s}^2\) on Earth, acting downwards. It causes any freely falling object to increase its velocity downward. In kinematics problems involving vertical movement, such as our flea's jump, this acceleration directly influences both the initial velocity necessary for an object to reach a certain height and the time it takes for the object to return to its initial position. The negative sign indicates the direction of acceleration is opposite to the upward direction of the initial jump.
Time of Flight
Time of flight refers to the total duration an object remains airborne. In the example of the flea jumping, the time of flight can be determined once we know the initial velocity and acceleration due to gravity. For an object projected vertically upwards, the time of flight is the time from launch to the moment it returns to the starting point. Calculating it uses the motion equation \[ v_f = v_i + a t, \] where we set the final velocity equal but opposite to the initial velocity when the object hits the ground again. By knowing the time of flight, we get a comprehensive view of how long the flea stays in the air during its jump.

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Most popular questions from this chapter

You are standing at rest at a bus stop. A bus moving at a constant speed of \(5.00 \mathrm{~m} / \mathrm{s}\) passes you. When the rear of the bus is \(12.0 \mathrm{~m}\) past you, you realize that it is your bus, so you start to run toward it with a constant acceleration of \(0.960 \mathrm{~m} / \mathrm{s}^{2}\). How far would you have to run before you catch up with the rear of the bus, and how fast must you be running then? Would an average college student be physically able to accomplish this?

Two stones are thrown vertically upward from the ground, one with three times the initial speed of the other. (a) If the faster stone takes \(10 \mathrm{~s}\) to return to the ground, how long will it take the slower stone to return? (b) If the slower stone reaches a maximum height of \(H,\) how high (in terms of \(H\) ) will the faster stone go? Assume free fall.

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 ball is thrown straight up from the ground with speed \(v_{0}\). At the same instant, a second ball is dropped from rest from a height \(H\) directly above the point where the first ball was thrown upward. There is no air resistance. (a) Find the time at which the two balls collide. (b) Find the value of \(H\) in terms of \(v_{0}\) and \(g\) such that at the instant when the balls collide, the first ball is at the highest point of its motion.

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?
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