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Chapter 2: Problem 39

(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
(a) Initial speed is approximately 2.94 m/s. (b) The flea is in the air for about 0.6 seconds.

Step by step solution

01

Identify the Known Values for Part (a)

We are given the maximum height of the flea's jump as 0.440 meters. We also know that the final velocity at the maximum height is 0 m/s and the acceleration is due to gravity, which is \( a = -9.8 \) m/s虏.
02

Apply the Kinematic Equation for Part (a)

Using the kinematic equation \( v^2 = u^2 + 2as \), we can solve for the initial velocity \( u \). Set final velocity \( v = 0 \), height \( s = 0.440 \) m, and acceleration \( a = -9.8 \) m/s虏. Rearrange to find \( u = \sqrt{v^2 - 2as} \).
03

Calculate Initial Speed for Part (a)

Substitute the values into the equation \( u = \sqrt{0 - 2(-9.8)(0.440)} \), resulting in \( u = \sqrt{8.624} \) which yields \( u \approx 2.94 \) m/s.
04

Understand the Motion for Part (b)

The total time of flight includes the time to ascend to the highest point and the time to descend back to the ground. The time to ascend equals the time to descend due to symmetrical motion under gravity.
05

Calculate Time to Reach Maximum Height for Part (b)

Use the equation \( v = u + at \), with \( v = 0 \), \( u = 2.94 \) m/s, and \( a = -9.8 \) m/s虏. Solve for \( t \) which gives \( 0 = 2.94 - 9.8t \), resulting in \( t = 0.3 \) seconds.
06

Determine Total Time in the Air for Part (b)

Since it takes 0.3 seconds to reach the peak, and due to symmetry, the descent takes another 0.3 seconds, the total air time is \( 0.3 + 0.3 = 0.6 \) seconds.

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Key Concepts

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

Projectile Motion
Projectile motion occurs when an object is launched into the air and is influenced by gravity, moving along a parabolic path. It involves both a vertical component, due to gravity, and sometimes potentially a horizontal component, if the object is thrown rather than simply dropped or shot vertically. In the case of vertical projectile motion, such as the jump of a flea, the object solely moves up and then back down in a straight line.

Key features of projectile motion include:
  • The trajectory shaped like a parabola for oblique projectiles, yet linear for purely vertical ones.
  • The presence of maximum height where the vertical velocity becomes zero before the object begins to descend.
  • The time of flight, which is the total time the projectile spends in the air.
Understanding projectile motion helps in analyzing various phenomena, such as calculating how long a basketball stays in the air or how high a flea can jump.
Gravity
Gravity plays a crucial role in determining the motion of a projectile. It is the force that pulls objects toward the Earth's surface, and in kinematic equations, it acts as a constant downward acceleration for objects close to Earth's surface. The magnitude of this force is approximately \(-9.8 \text{ m/s}^2\), indicating that gravity decelerates upward motion and accelerates downward motion.

Gravity affects:
  • The maximum height an object can achieve, as it decelerates the object during ascent until its vertical velocity reaches zero.
  • The time of flight, because an object takes equal time to reach peak altitude as it does to descend back under symmetrical conditions.
When an object is launched upward, gravity works against its motion, eventually stopping its ascent entirely and pulling it back to the ground.
Initial Velocity Calculation
Calculating the initial velocity is an important step to solve problems involving projectile motion. The initial velocity refers to the speed at which an object is launched upwards or thrown. In the context of kinematic equations, initial velocity is denoted as \( u \).

To calculate the initial velocity for an upward projectile motion, you can use the kinematic formula \( v^2 = u^2 + 2as \), where:
  • \( v \): final velocity at the highest point (0 m/s for a vertical ascent).
  • \( u \): initial velocity (what you aim to find).
  • \( a \): acceleration due to gravity (\(-9.8 \text{ m/s}^2\)).
  • \( s \): vertical displacement or maximum height (0.440 m in the flea's case).
By rearranging the equation and inserting known values, you solve for \( u \), resulting in the initial speed necessary for reaching a given height or other specified trajectory path.

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Most popular questions from this chapter

Earthquake Analysis. Earthquakes produce several types of shock waves. The most well known are the P-waves (P for primary or pressure) and the S-waves (S for secondary or shear). In the earth's crust, the P-waves travel at around 6.5 \(\mathrm{km} / \mathrm{s}\) , while the S-waves move at about 3.5 \(\mathrm{km} / \mathrm{s}\) . The actual speeds vary depending on the type of material they are going through. The time delay between the arrival of these two waves at a seismic recording station tells geologists how far away the earthquake occurred. If the time delay is 33 \(\mathrm{s}\) , how far from the seismic station did the earthquake occur?

The acceleration of a bus is given by \(a_{x}(t)=\alpha t,\) where \(\alpha=1.2 \mathrm{m} / \mathrm{s}^{3} .\) (a) If the bus's velocity at time \(t=1.0 \mathrm{s}\) is 5.0 \(\mathrm{m} / \mathrm{s}\) , what is its velocity at time \(t=2.0 \mathrm{s} ?(\mathrm{b})\) If the bus's position at time \(t=1.0 \mathrm{s}\) is \(6.0 \mathrm{m},\) what is its position at time \(t=2.0 \mathrm{s} ?\) (c) Sketch \(a_{x}-t, v_{x}-t,\) and \(x-\) graphs for the motion.

The driver of a car wishes to pass a truck that is traveling at a constant speed of 20.0 \(\mathrm{m} / \mathrm{s}\) (about 45 \(\mathrm{mi} / \mathrm{h} )\) . Initially, the car is also traveling at 20.0 \(\mathrm{m} / \mathrm{s}\) and its front bumper is 24.0 \(\mathrm{m}\) behind the truck's rear bumper. The car accelerates at a constant 0.600 \(\mathrm{m} / \mathrm{s}^{2}\) , then pulls back into the truck's lane when the rear of the car is 26.0 \(\mathrm{m}\) ahead of the front of the truck. The car is 4.5 \(\mathrm{m}\) long and the truck is 21.0 \(\mathrm{m}\) long. (a) How much time is required for the car to pass the truck? (b) What distance does the car travel during this time? (c) What is the final speed of the car?

A turtle crawls along a straight line, which we will call the \(x\) -axis with the positive dircction to the right. The equation for the turtle's position as a function of time is \(x(t)=50.0 \mathrm{cm}+\) \((2.00 \mathrm{cm} / \mathrm{s}) t-\left(0.0625 \mathrm{cm} / \mathrm{s}^{2}\right) t^{2},(\mathrm{a})\) Find the turte'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 these times? (e) Sketch graphs of \(x\) versus \(t, v_{x}\) versus \(t,\) and \(a_{x}\) versus \(t\) for the time interval \(t=0\) to \(t=40 \mathrm{s}\)

Two sunt drivers drive directly toward each other. At time \(t=0\) the two cars are a distance \(D\) apart, car 1 is at rest, and \(\operatorname{car} 2\) is moving to the left with speed \(v_{0}\) . Car 1 begins to move at \(t=0\) , speeding up with a constant acceleration \(a_{x}\) Car 2 continues to move with a constant velocity. (a) At what time do the two cars collide? \((b)\) Find the speed of car 1 just before it collides with \(\operatorname{car} 2 .\) (c) Sketch \(x-t\) and \(v_{x}-t\) graphs for car 1 and \(\operatorname{car} 2 .\) For each of the two graphs, draw the curves for both cars on the same set of axes.
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