/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 Two balloons are simultaneously ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 19

Two balloons are simultaneously released from two buildings \(A\) and \(B\). Balloon from A rises with constant velocity \(10 \mathrm{~ms}^{-1}\), While the other one rises with constant velocity of \(20 \mathrm{~ms}^{-1}\). Due to wind the balloons gather horizontal velocity \(V_{x}=0.5 y\), where \(y\) is the height from the point of release. The buildings are at a distance of \(250 \mathrm{~m}\) and after some time \(t\) the balloons collide. Choose the correct option(s). (1) \(t=5 \mathrm{sec}\) (2) difference in height of buildings is \(100 \mathrm{~m}\) (3) difference in height of buildings is \(50 \mathrm{~m}\) (4) \(t=10 \mathrm{sec}\)

Short Answer

Expert verified
The correct options are (2) difference in height of buildings is 100 m and (4) \( t = 10 \) sec.

Step by step solution

01

Calculate Balloon Heights

The height of the balloon released from building \(A\) is given by \( y_A = v_A \cdot t = 10t \). The height of the balloon released from building \(B\) is given by \( y_B = v_B \cdot t = 20t \).
02

Calculate Horizontal Displacements

Integrate the horizontal velocity function to find horizontal displacements. For balloon \(A\), the horizontal velocity is \( V_{x_A} = 0.5 \cdot y_A = 0.5 \cdot 10t = 5t \). Its horizontal displacement is \( x_A = \int 5t \, dt = \frac{5}{2} t^2 \). Similarly, for balloon \(B\), \( V_{x_B} = 0.5 \cdot y_B = 0.5 \cdot 20t = 10t \) and its displacement is \( x_B = \int 10t \, dt = 5t^2 \).
03

Distance Between Balloons

The horizontal distance between the balloons when they meet is the distance between the buildings: \( x_B - x_A = 250 \). Putting the displacement equations, we get \( 5t^2 - \frac{5}{2} t^2 = 250 \). Simplifying, \( \frac{5}{2} t^2 = 250 \).
04

Solve for Time (t)

Solve the equation \( \frac{5}{2} t^2 = 250 \) to find \( t \). First, divide both sides by \( \frac{5}{2} \): \( t^2 = \frac{250 \times 2}{5} = 100 \). Then, take the square root: \( t = 10 \) seconds.
05

Calculate Height Difference Between Buildings

The difference in the heights of the buildings from which the balloons started is the initial height difference when \( t = 0 \). Hence,\( |y_B - y_A| \) at the time of impact when the balloons have risen \( t = 10 \) secs gives the difference in height. \( |y_B - y_A| = |20(10) - 10(10)| = 100 \, \text{m} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Horizontal Velocity
In projectile motion, horizontal velocity is an essential aspect, especially when wind or another force affects the movement of an object. Here, the balloons acquire a horizontal velocity due to wind, expressed as \( V_{x} = 0.5y \). This means that the horizontal speed is directly affected by the height \( y \) from the release point. As the height of the balloons increases, their horizontal velocity increases as well. This is because the function \( V_{x} = 0.5y \) demonstrates a linear relationship:
  • At any height \( y \), the horizontal velocity \( V_{x} \) is simply \( 0.5 \) times that height.
  • This allows us to calculate how far the balloons travel horizontally as they rise.
Understanding horizontal velocity helps us compute how fast the balloons move away from or toward each other and is integral to determining their meeting point.
Constant Acceleration
In projectile motion terms, constant acceleration often refers to acceleration due to gravity. However, in this exercise, the focus is instead on the constant acceleration of the vertical movement of the balloons. Each balloon rises with a constant velocity, implying a uniform rate of height increase:
  • Balloon from building A ascends at \( 10 \text{ ms}^{-1} \).
  • Balloon from building B ascends faster at \( 20 \text{ ms}^{-1} \).
Though the velocities are constant, they effectively cause a constant 'acceleration' of vertical position over time. This steady pace ensures we can straightforwardly calculate the height of each balloon at any given time \( t \), an essential step when assessing their paths and meeting conditions.
Collision Time
Collision time is the moment when both balloons reach the same horizontal position, leading to their paths crossing. By setting the difference between their horizontal positions equal to the distance between the buildings (250 meters), we calculate the collision time:
  • The task is to solve the equation \( x_B - x_A = 250 \).
  • Each balloon鈥檚 horizontal motion keeps pace with their rising height thanks to wind-induced velocity.
Utilizing the expressions for horizontal displacement \( x_A = \frac{5}{2} t^2 \) and \( x_B = 5t^2 \), plugging them into the equation gives us the sole collision time of \( t = 10 \) seconds. Knowing exactly when and where these paths intersect is crucial to analyzing projectile motion scenarios.
Height Difference
The height difference at the moment of collision between balloons is vital to understanding the disparity in initial heights of their respective buildings. When we speak of a height difference in this context, it鈥檚 about finding the disparity in elevation that led them to collude while their vertical ascent continues:
  • The formula for each balloon鈥檚 height at the time \( t \) is \( y_A = 10t \) and \( y_B = 20t \).
  • At collision time \( t = 10 \text{ s} \), their respective heights are 100 meters and 200 meters.
As a result, the initial height difference between the buildings, as discerned from the risen heights, is \( |y_B - y_A| = 100 \, \text{m} \). This calculation helps explain the feasible scenarios under which such a collision can occur.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A ball is thrown at different angles with the same speed \(u\) and from the same point and it has the same range in both the cases. If \(y_{1}\) and \(y_{2}\) are the heights attained in the two cases, then \(y_{1}+y_{2}\) is equal to (1) \(\frac{u^{2}}{g}\) (2) \(\frac{2 u^{2}}{g}\) (3) \(\frac{u^{2}}{2 g}\) (4) \(\frac{u^{2}}{4 g}\)

A swimmer wishes to cross a \(500-\mathrm{m}\) river flowing at \(5 \mathrm{~km} \mathrm{~h}^{-1}\). His speed with respect to water is \(3 \mathrm{~km} \mathrm{~h}^{-1}\). The shortest possible time to cross the river is (1) \(10 \mathrm{~min}\) (2) \(20 \mathrm{~min}\) (3) \(6 \mathrm{~min}\) (4) \(7.5 \mathrm{~min}\)

A river flows with a speed more than the maximum speed with which a person can swim in still water. He intends to cross the river by the shortest possible path (i.e., he wants to reach the point on the opposite bank which directly opposite to the starting point). Which of the following is correct? (1) He should start normal to the river bank. (2) He should start in such a way that he moves normal to the bank, relative to the bank. (3) He should start in a particular (calculated) direction making an obtuse angle with the direction of water current. (4) The man cannot cross the river in that way.

An aeroplane is flying vertically upwards. When it is at a height of \(1000 \mathrm{~m}\) above the ground and moving at a speed of \(367 \mathrm{~m} / \mathrm{s}\), a shot is fired at it with a speed of \(567 \mathrm{~m} \mathrm{~s}^{-1}\) from a point directly below it. What should be the acceleration of aeroplane so that it may escape from being hit? (1) \(>5 \mathrm{~ms}^{-2}\) (2) \(>10 \mathrm{~ms}^{-2}\) (3) \(<10 \mathrm{~ms}^{-2}\) (4) Not possible

Rain appears to fall vertically to a man walking at \(3 \mathrm{~km} \mathrm{~h}^{-1}\), but when he changes his speed to double, the rain appears to fall at \(45^{\circ}\) with vertical. Study the following statements and find which of them are correct. i. Velocity of rain is \(2 \sqrt{3} \mathrm{~km} \mathrm{~h}^{-1}\) ii. The angle of fall of rain (with vertical) is $$ \theta=\tan ^{-1}\left(\frac{1}{\sqrt{2}}\right) \text {. } $$ iii. The angle of fall of rain (with vertical) is $$ \theta=\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right) \text {. } $$ iv. Velocity of rain is \(3 \sqrt{2} \mathrm{kmh}^{-1}\). (I) Statements (i) and (ii) are correct. (2) Statements (i) and (iii) are correct. (3) Statements (iii) and (iv) are correct. (4) Statements (ii) and (iv) are correct.
See all solutions

Recommended explanations on Physics Textbooks

Electrostatics

Read Explanation

Nuclear Physics

Read Explanation

Energy Physics

Read Explanation

Classical Mechanics

Read Explanation

Electricity and Magnetism

Read Explanation

Kinematics Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.