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Chapter 2: Problem 59

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 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.

Short Answer

Expert verified
(a) Slower stone takes approximately 6.67 seconds. (b) Faster stone reaches 9H.

Step by step solution

01

Understand Problem Statement and Gather Information

The problem involves two stones thrown vertically, each with different initial velocities. The faster stone's initial speed is three times that of the slower stone. The faster stone takes 10 seconds to return to the ground.
02

Apply Kinematics Equation for Time of Flight

For vertical displacement under constant acceleration, the time taken for an object to return to its original height when thrown upwards is given by: \( t = \frac{2v_0}{g} \). Given that the faster stone takes 10 s, we can say \( 10 = \frac{2(3v_0)}{g} \). Solve for \( v_0 \) first.
03

Solve for Initial Speed of Slower Stone

From the equation \( 10 = \frac{6v_0}{g} \), rearrange to find \( v_0 = \frac{10g}{6} = \frac{5g}{3} \).
04

Calculate Time for the Slower Stone's Return

For the slower stone, its total time of flight is \( t = \frac{2v_0}{g} = \frac{2(\frac{5g}{3})}{g} = \frac{10}{3} \), solving for \( t \), the time of flight for the slower stone is \( \frac{10}{3} \times 2 = \frac{20}{3} \approx 6.67 \) seconds.
05

Determine Maximum Height in Terms of H

The maximum height reached by a projectile is given by \( h = \frac{v_0^2}{2g} \). If the slower stone reaches a maximum height \( H \), \( H = \frac{v_0^2}{2g} \). For the faster stone: it reaches a height \( \frac{(3v_0)^2}{2g} = \frac{9v_0^2}{2g} \). This is 9 times the height of the slower stone. Thus, in terms of \( H \), the height is \( 9H \).

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Key Concepts

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

Kinematics Equations
Kinematics equations are essential tools in physics that help us understand the motion of objects. These equations relate the five key variables of motion: displacement, initial velocity, final velocity, acceleration, and time. For vertical motion like throwing stones upwards, these equations simplify the problem.
  • The primary equation often used is: \( v = v_0 + at \)
  • Displacement can be calculated using: \( s = v_0t + \frac{1}{2}at^2 \)
  • When dealing with maximum height or return time, \( v^2 = v_0^2 + 2as \) is useful.

In the provided exercise, the kinematics equation for time of flight and maximum height are utilized. They directly relate initial velocity, time, and the gravitational acceleration \( g \). These relations make it possible to solve problems where direct measurements of height and time are not possible.
Projectile Motion
In physics, projectile motion deals with objects that are thrown or propelled into the air, subject to gravitational forces. Vertical motion is a straightforward type of projectile motion. Here, the object only moves up and down.
A key feature of vertical projectile motion is that the only force acting on the projectile after it is thrown is gravity. This means:
  • Acceleration is constant and equals \(-g\) (approximately \(-9.81 \text{ ms}^{-2}\)).
  • At maximum height, the velocity of the projectile is zero.
  • The time taken to rise equals the time taken to descend back to the point of throw.
Understanding vertical motion helps to tackle problems where objects move in arcs or are simply dropped or thrown upwards.
Time of Flight
Time of flight is the total time an object spends in the air. For objects projected vertically, we can calculate this using the formula \( t = \frac{2v_0}{g} \). Here, \( v_0 \) represents initial velocity and \( g \) is the acceleration due to gravity.
This formula derives from the fact that the time to reach maximum height equals the time to fall back down. The exercise illustrates using this formula to calculate the time taken by two stones differently propelled. If one stone takes 10 seconds, the time for the slower stone would be computed given a third of the initial velocity.
  • Time of flight is symmetrical in vertical motion.
  • The same formula can apply showing proportionality, such as a faster stone taking longer.
Maximum Height
When discussing projectile motion, maximum height is the highest point reached by the object. For vertically projected objects, use the formula \( h = \frac{v_0^2}{2g} \) where \( v_0 \) is the initial velocity and \( g \) is gravity's acceleration.
In the exercise, we find the height in terms of \( H \) for two stones. The faster one, with an initial speed triple that of the slower, attains a height nine times greater because both velocity and height scale with the square of the velocity ratio.
  • At maximum height, the object鈥檚 velocity is zero.
  • Height increases as the square of velocity doubles.
Understanding maximum height explains why projectiles might achieve different altitudes when thrown at varying efforts.

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Most popular questions from this chapter

A subway train starts from rest at a station and accelerates at a rate of 1.60 \(\mathrm{m} / \mathrm{s}^{2}\) for 14.0 s. It runs at constant speed for 70.0 \(\mathrm{s}\) and slows down at a rate of 3.50 \(\mathrm{m} / \mathrm{s}^{2}\) until it stops at the next station. Find the total distance covered.

Egg drop. You are on the roof of the physics building of your school, 46.0 \(\mathrm{m}\) above the ground. (See Figure \(2.53 . )\) Your physics professor, who is 1.80 \(\mathrm{m}\) tall, is walking alongside the building at a constant speed of 1.20 \(\mathrm{m} / \mathrm{s} .\) If you wish to drop an egg on your professor's head, where should the professor be when you release the egg, assuming that the egg encounters no appreciable air drag.

Air-bag injuries. During an auto accident, the vehicle's air bags deploy and slow down the passengers more gently than if they had hit the windshield or steering wheel. According to safety standards, the bags produce a maximum acceleration of \(60 \mathrm{g},\) but lasting for only 36 \(\mathrm{ms}\) (or less). How far (in meters) does a person travel in coming to a complete stop in 36 \(\mathrm{ms}\) at a constant acceleration of 60 \(\mathrm{g}\) ?

The wind is blowing from west to east at 35 \(\mathrm{mph}\) , and an eagle in that wind is flying at 22 mph relative to the air. What is the velocity of this eagle relative to a person standing on the ground if the eagle is flying (a) from west to east relative to the air and (b) from east to west relative to the air?

In an experiment, a shearwater (a seabird) was taken from its nest, flown 5150 \(\mathrm{km}\) away, and released. The bird found its way back to its nest 13.5 days after release. If we place the origin in the nest and extend the \(+x\) -axis to the release point, what was the bird's average velocity in \(\mathrm{m} / \mathrm{s}\) (a) for the return flight, and (b) for the whole episode, from leaving the nest to returning?
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