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Chapter 3: Problem 14

The froghopper, \(Philaenus\) \(spumarius\), holds the world record for insect jumps. When leaping at an angle of 58.0\(^\circ\) above the horizontal, some of the tiny critters have reached a maximum height of 58.7 cm above the level ground. (See \(Nature\), Vol. 424, July 31, 2003, p. 509.) (a) What was the takeoff speed for such a leap? (b) What horizontal distance did the froghopper cover for this world-record leap?

Short Answer

Expert verified
Takeoff speed: 3.94 m/s; Horizontal distance: 1.33 m.

Step by step solution

01

Identify Given Values

We are given the leap angle \( \theta = 58.0^\circ \) and the maximum height \( h = 58.7 \text{ cm} = 0.587 \text{ m} \). We need to find the takeoff speed \( v_0 \) and the horizontal distance covered.
02

Calculation of Takeoff Speed

Using the formula for maximum height in projectile motion, \( h = \frac{v_{0y}^2}{2g} \), where \( v_{0y} = v_0 \sin \theta \) and \( g = 9.81 \text{ m/s}^2 \). Rearrange to solve for \( v_0 \):\[v_0 = \sqrt{\frac{2gh}{\sin^2 \theta}}.\]Substitute \( g = 9.81 \text{ m/s}^2 \), \( h = 0.587 \text{ m} \), and \( \theta = 58.0^\circ \):\[v_0 = \sqrt{\frac{2 \times 9.81 \times 0.587}{\sin^2(58.0^\circ)}} \approx 3.94 \text{ m/s}.\]
03

Determine Horizontal Component of Velocity

The horizontal component of the takeoff speed is \( v_{0x} = v_0 \cos \theta \). Using \( v_0 = 3.94 \text{ m/s} \) and \( \theta = 58.0^\circ \):\[v_{0x} = 3.94 \cos(58.0^\circ) \approx 2.08 \text{ m/s}.\]
04

Find Time of Flight

The time to reach maximum height is given by \( t_{up} = \frac{v_{0y}}{g} \), where \( v_{0y} = v_0 \sin \theta \). Therefore,\[t_{up} = \frac{3.94 \sin(58.0^\circ)}{9.81} \approx 0.32 \text{ s}.\]Since the flight is symmetric, the total time of flight \( T = 2 \times t_{up} \approx 0.64 \text{ s}.\)
05

Calculate Horizontal Distance Covered

The horizontal distance, or range, \( x \) is given by \( x = v_{0x} \times T \). Substitute \( v_{0x} = 2.08 \text{ m/s} \) and \( T = 0.64 \text{ s} \):\[x = 2.08 \times 0.64 \approx 1.33 \text{ m}.\]

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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 mechanics that deals with the motion of objects. It does not concern itself with the forces that cause this motion. In this exercise, we explore a specific case of motion, known as projectile motion. The froghopper's leap is a fascinating example of how we can use kinematics to describe complex movements. We divide the motion into vertical and horizontal components, allowing us to study their behavior individually.
  • The vertical motion is influenced by gravity, and we use the maximum height formula to understand it. This is where the froghopper's takeoff speed becomes crucial, as it tells us how fast it must launch to reach a certain height.
  • The horizontal motion remains constant during the leap, unaffected by gravity. Here, we use kinematics to determine how far the froghopper travels horizontally. By analyzing both components, we get a comprehensive picture of the leap's dynamics.
Understanding kinematics is essential in physics, as it provides the tools to analyze motion, not only for small insects but also for larger objects in the real world.
Physics Problems
Physics problems often involve scenarios where various physical quantities interact. They teach us to apply theoretical concepts to practical situations. In solving the froghopper's leap, we first identify given values like angle and height. This helps set up the problem effectively.
  • Next, we utilize equations of motion to calculate desired quantities, such as the takeoff speed and horizontal distance. These calculations require a careful understanding of how different concepts like angles, velocities, and time of flight interrelate.
  • Following a structured approach helps in breaking down complex problems into manageable steps. By treating each step methodically, you can derive the desired results without rushing or making errors.
Physics problems enhance problem-solving skills and help develop logical thinking, both of which are crucial in scientific studies and practical applications.
Two-dimensional Motion
Two-dimensional motion occurs when an object moves in a plane, having both a horizontal and a vertical component. It is crucial when studying projectiles, like the froghopper's record leap.
  • This type of motion involves separating the overall movement into independent vertical and horizontal motions. Each component is analyzed separately, using its own set of equations related to time, velocity, and displacement.
  • For vertical motion governed by gravity, we use the sine component of the angle to determine initial vertical velocity, helping us find the maximum height. In contrast, the horizontal component relies on the cosine to ascertain how far the froghopper will travel from its starting point.
Understanding two-dimensional motion helps us predict and calculate the movement paths of various objects, from tiny insects to large-segment movements in mechanics. It's an essential concept in physics that broadens our comprehension of how objects interact with their environment.

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Most popular questions from this chapter

The radius of the earth's orbit around the sun (assumed to be circular) is 1.50 \(\times\) 10\(^8\) km, and the earth travels around this orbit in 365 days. (a) What is the magnitude of the orbital velocity of the earth, in m/s? (b) What is the radial acceleration of the earth toward the sun, in m/s\(^2\) ? (c) Repeat parts (a) and (b) for the motion of the planet Mercury (orbit radius = 5.79 \(\times\) 10\(^7\) km, orbital period = 88.0 days).

Firemen use a high-pressure hose to shoot a stream of water at a burning building. The water has a speed of 25.0 m/s as it leaves the end of the hose and then exhibits projectile motion. The firemen adjust the angle of elevation \(\alpha\) of the hose until the water takes 3.00 s to reach a building 45.0 m away. Ignore air resistance; assume that the end of the hose is at ground level. (a) Find \(\alpha\). (b) Find the speed and acceleration of the water at the highest point in its trajectory. (c) How high above the ground does the water strike the building, and how fast is it moving just before it hits the building?

A boy 12.0 m above the ground in a tree throws a ball for his dog, who is standing right below the tree and starts running the instant the ball is thrown. If the boy throws the ball horizontally at 8.50 m/s, (a) how fast must the dog run to catch the ball just as it reaches the ground, and (b) how far from the tree will the dog catch the ball?

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During a storm, a car traveling on a level horizontal road comes upon a bridge that has washed out. The driver must get to the other side, so he decides to try leaping the river with his car. The side of the road the car is on is 21.3 m above the river, while the opposite side is only 1.8 m above the river. The river itself is a raging torrent 48.0 m wide. (a) How fast should the car be traveling at the time it leaves the road in order just to clear the river and land safely on the opposite side? (b) What is the speed of the car just before it lands on the other side?
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