/*! 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 44 A hot-air balloonist, rising ver... [FREE SOLUTION] | 91Ó°ÊÓ

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A hot-air balloonist, rising vertically with a constant velocity of magnitude \(5.00 \mathrm{m} / \mathrm{s},\) releases a sandbag at an instant when the balloon is 40.0 \(\mathrm{m}\) above the ground (Fig. 2.41\()\) . After it is released, the sandbag is in frec fall. (a) Compute the position and velocity of the sandbag at 0.250 s and 1.00 s after its release. (b) How many seconds s strike after its release will the bag strike the ground? (c) With what magnitude of velocity does it strike the ground? (d) What is the greatest height above the ground that the sandbag reaches? (e) Sketch \(a_{y}-t\) \(v_{y^{-} t},\) and \(y-t\) graphs for the motion.

Short Answer

Expert verified
Compute position/velocity at given times, find time to hit the ground, velocity at impact, and max height. Sketch graphs for motion.

Step by step solution

01

Initial Conditions

When the sandbag is released, its initial position is 40.0 m above the ground, and it has the same vertical velocity as the balloon of 5.00 m/s upwards. The acceleration acting on the sandbag is gravity, which is -9.81 m/s² (acting downwards).
02

Position at 0.25 Seconds

Use the equation for position in uniformly accelerated motion: \[ y = y_0 + v_0 t + \frac{1}{2} a t^2 \]Substitute: \( y_0 = 40.0 \) m, \( v_0 = 5.00 \) m/s, \( a = -9.81 \) m/s², and \( t = 0.25 \) s: \[ y = 40.0 + 5.00 \times 0.25 + \frac{1}{2} (-9.81) \times (0.25)^2 \] Calculate to find \( y \) position.
03

Velocity at 0.25 Seconds

Use the equation for velocity in uniformly accelerated motion: \[ v = v_0 + a t \]Substitute: \( v_0 = 5.00 \) m/s, \( a = -9.81 \) m/s², and \( t = 0.25 \) s: \[ v = 5.00 + (-9.81) \times 0.25 \] Calculate to find velocity \( v \).
04

Position at 1.0 Second

Using the same position equation: \[ y = y_0 + v_0 t + \frac{1}{2} a t^2 \]Substitute: \( y_0 = 40.0 \) m, \( v_0 = 5.00 \) m/s, \( a = -9.81 \) m/s², and \( t = 1.0 \) s: \[ y = 40.0 + 5.00 \times 1.0 + \frac{1}{2} (-9.81) \times (1.0)^2 \] Calculate to find \( y \) position.
05

Velocity at 1.0 Second

Using the same velocity equation: \[ v = v_0 + a t \]Substitute: \( v_0 = 5.00 \) m/s, \( a = -9.81 \) m/s², and \( t = 1.0 \) s: \[ v = 5.00 + (-9.81) \times 1.0 \] Calculate to find velocity \( v \).
06

Time to Hit Ground

Use equation of motion where final position is 0 (ground level):\[ 0 = y_0 + v_0 t + \frac{1}{2} a t^2 \]Substitute \( y_0 = 40.0 \), \( v_0 = 5.00 \), \( a = -9.81 \), and solve for \( t \). This is a quadratic equation and can be solved using the quadratic formula.
07

Velocity Impact

Once time \( t \) to hit the ground has been found, substitute this \( t \) back into the velocity equation \( v = v_0 + a t \) to determine the velocity at impact.
08

Maximum Height

The maximum height is when the velocity changes direction, i.e., when \( v = 0 \). Use \( 0 = v_0 + a t \) to solve for \( t \), find the time at which the maximum height occurs, and substitute this \( t \) back into the position equation to find maximum height.
09

Graph Sketching

Create three graphs: (1) Acceleration vs Time \(a_y-t\),which is a constant -9.81 m/s². (2) Velocity vs Time \(v_y-t\), starting at 5.00 m/s and decreasing linearly. (3) Position vs Time \(y-t\), a parabolic trajectory peaking at the maximum height.

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Key Concepts

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

Uniformly Accelerated Motion
Understanding projectile motion starts with mastering uniformly accelerated motion. This describes the motion of objects moving under a constant acceleration, like gravity. In our scenario, after the release of the sandbag from the hot-air balloon, it starts to move downwards due to the acceleration from gravity, which is constant and has a magnitude of approximately \(-9.81 \text{ m/s}^2\). It's essential to note that this acceleration acts in one direction - downwards.With this uniform acceleration, the position and velocity of the sandbag can be calculated at any given time using kinematic equations. The process starts with identifying initial conditions, such as the initial position (e.g., \(40.0 \text{ m}\) above ground) and initial velocity (the same as the balloon's at the point of release: \(5.00 \text{ m/s}\) upwards). Upon release, the sandbag follows a predictable path, influenced directly by gravity.
Kinematics Equations
The key to analyzing uniformly accelerated motion lies in using kinematics equations. These equations allow you to determine the position and velocity of an object when its acceleration is constant. For this problem, we primarily use two:
  • Position: \( y = y_0 + v_0 \cdot t + \frac{1}{2} a t^2 \)
  • Velocity: \( v = v_0 + a \cdot t \)
Each term in these equations represents specific aspects of the motion:
  • \(y_0\) is the initial position, \(v_0\) is the initial velocity.
  • \(a\) is the acceleration.
  • \(t\) is the time elapsed since the start of the motion.
By substituting the known values into these equations, you can solve for unknown quantities. For example, calculating the sandbag's position and velocity at 0.25 seconds required substituting \(y_0 = 40.0 \text{ m}\), \(v_0 = 5.00 \text{ m/s}\), and \(a = -9.81 \text{ m/s}^2\). This method illustrates how the application of kinematics equations is crucial for analyzing projectile motion.
Vertical Velocity
The vertical velocity of an object in uniformly accelerated motion, such as the sandbag, changes linearly over time due to gravity. Initial vertical velocity is crucial in determining how quickly and in which direction the object moves initially. In our problem, the sandbag's initial vertical velocity is equal to that of the hot-air balloon at the moment of release \(5.00 \text{ m/s}\), but in an upward direction.Over time, the effect of gravity causes this velocity to decrease as it moves upwards until it momentarily becomes zero at the peak of its trajectory. After reaching this peak, the sandbag starts to fall back down, and the vertical velocity increases again in the downward direction. Calculating the vertical velocity at different times like 0.25 s or 1.0 s gives insights into how the speed changes at different points during the fall.
Acceleration due to Gravity
Gravity is a constant downward acceleration affecting objects near the earth's surface, approximately \(9.81 \text{ m/s}^2\). In projectile motion, this constant force accelerates the object downward, regardless of its initial upward velocity. It's helpful to remember:
  • Gravity's effect begins immediately upon the sandbag's release.
  • This force remains constant and points towards the earth's center.
  • It determines how quickly an object speeds up or slows down vertically.
For the sandbag, gravity acts as a decelerating force when moving upwards and an accelerating force on its way down. With gravity as the sole force in flight (assuming air resistance is negligible), it uniformly accelerates the sandbag downwards. Understanding this concept helps in predicting and calculating the motion path using the equations provided.

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Most popular questions from this chapter

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