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Chapter 2: Problem 94

An apple drops from the tree and falls frecly. The apple is originally at rest a height \(H\) above the top of the grass of a thick lawn, which is made of blades of grass of height \(h\) . When the apple enters the grass, it slows down at a constant rate so that its speed is 0 when it reaches ground level. (a) Find the speed of the apple just before it enters the grass. (b) Find the acceleration of the apple while it is in the grass. (c) Sketch the \(y-t, v_{y}-t,\) and \(a_{y}-t\) graphs for the apple's motion.

Short Answer

Expert verified
(a) Speed is \( \sqrt{2gH} \), (b) acceleration is \\(-\frac{(2gH)}{2h} \\).

Step by step solution

01

Understanding the Problem

We need to examine the motion of an apple dropping from a height. Initially at rest, the apple will accelerate due to gravity and eventually pass through the grass where it decelerates to a stop at the ground. We will find the speed just before it hits the grass, the acceleration in the grass, and sketch motion graphs.
02

Calculate speed before entering grass

To find the speed just before the apple enters the grass, we use the formula for an object in free fall: \[ v = \sqrt{2gH} \] where \( g \) is the acceleration due to gravity (approximately \( 9.81 \ m/s^2 \)). This formula considers the potential energy at height \( H \) converting entirely into kinetic energy just before reaching the grass.
03

Calculate acceleration in the grass

The apple decelerates uniformly in the grass. We use the kinematic equation: \[ v^2 = u^2 + 2as \] where \( u \) is the initial velocity (speed before entering the grass), \( v \) is the final velocity (0, when it stops), \( a \) is the acceleration, and \( s \) is the distance through the grass (\( h \)). Solving for \( a \), we have: \[ a = -\frac{u^2}{2h} \] Note the negative sign, indicating deceleration.
04

Sketch motion graphs

1. **Position vs. Time graph (\( y-t \))**: Starts at \( H \), curves down exponentially until \( H-h \), then linearly reaches 0.2. **Velocity vs. Time graph (\( v_{y}-t \))**: Linearly decreases from initial speed \( \sqrt{2gH} \) to 0.3. **Acceleration vs. Time graph (\( a_{y}-t \))**: Constant \( -g \) during free-fall, then constant negative value in the grass.

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Key Concepts

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

Free Fall
When an object is in free fall, it moves under the influence of gravity alone. In the case of the apple, it starts from rest and accelerates towards the ground. This acceleration is due to gravity, which has a constant value around the Earth's surface, approximately 9.81 m/s虏. During free fall, the only force acting on the apple is gravity, causing it to gain speed.

This acceleration continues until it hits the thick lawn grass. Here, potential energy at the starting height (just before falling) is completely converted into kinetic energy just before the apple enters the grass. This is described by the formula: \( v = \sqrt{2gH} \).

Key points about free fall:
  • Only affected by gravity.
  • Keeps accelerating while in the air.
  • No other forces considered during this phase.
Constant Acceleration
During both free fall and motion through the grass, the apple experiences constant acceleration, though of different magnitudes. In free fall, the acceleration is due to gravity, remaining consistently at 9.81 m/s虏.

However, once the apple enters the grass, it undergoes a constant deceleration. This steady decrease in speed is characterized by a constant acceleration value which is negative, signifying slowing down. It can be calculated using the rearranged kinematic equation: \[ a = -\frac{u^2}{2h} \] where \( u \) is the speed just before entering the grass, and \( h \) is the thickness of the grass.

Understanding constant acceleration:
  • Gravity causes a constant increase in velocity during free fall.
  • Grass applies a uniform resistive force, causing constant slowing.
  • Sign of acceleration indicates direction (negative for deceleration).
Velocity-Time Graph
The velocity-time graph visualizes how the apple's velocity changes during its fall and as it passes through the grass. Initially, the velocity increases linearly with time during the free fall due to gravity, reflecting constant acceleration. This portion of the graph is a straight line with a positive slope.

Once the apple starts moving through the grass, the graph depicts a linear decline from the maximum speed, indicating constant deceleration. Eventually, the line reaches zero as the apple completely stops at the earth's surface.

Important aspects of the velocity-time graph:
  • Positive slope during free fall (constant acceleration of gravity).
  • Negative slope during deceleration (constant deceleration in grass).
  • Area under the curve represents change in position (distance traveled).
Acceleration-Time Graph
An acceleration-time graph shows how acceleration varies with time during the apple's motion. For the apple's fall, the graph starts with a horizontal line at -9.81 m/s虏, indicating constant gravitational acceleration downward.

When the apple enters the grass, the acceleration suddenly changes to a different constant negative value calculated by \( a = -\frac{u^2}{2h} \). This reflects the deceleration due to the resistance provided by the grass until the apple reaches a stop.

Key details on the acceleration-time graph:
  • Displays constant negative values during both free fall and deceleration.
  • Helps understand force applied on the apple (gravity and grass resistance).
  • Sharp change when entering the grass corresponds to change in acceleration type.

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Most popular questions from this chapter

A marble is released from one rim of a hemispherical bowl of diameter 50.0 \(\mathrm{cm}\) and rolls down and up to the opposite rim in 10.0 s. Find (a) the uverage speed and (b) the average velocity of the marble.

Dan gets on Interstate Highway \(\mathrm{I}-80\) at Seward, Nebraska, and drives due west in a straight line and at an average velocity of magnitude 88 \(\mathrm{km} / \mathrm{h}\) . After traveling 76 \(\mathrm{km}\) , he reaches the Aurora exit (Fig. 2.44\()\) . Realizing he has gone too far, he turns around and drives due east 34 \(\mathrm{km}\) back to the York exit at an average velocity of magnitude 72 \(\mathrm{km} / \mathrm{h}\) . For his whole trip from Seward to the York exit, what are (a) his average speed and (b) the magnitude of his average velocity?

A world-class sprinter accelerates to his maximum speed in 4.0 \(\mathrm{s}\) . He then maintains this speed for the remainder of a \(100-\mathrm{m}\) race, finishing with a total time of 9.1 \(\mathrm{s}\) . (a) What is the runner's average acceleration during the first 4.0 \(\mathrm{s} ?(\mathrm{b})\) What is his average acceleration during the last 5.1 \(\mathrm{s} ?(\mathrm{c})\) What is his average acceleration for the entire race? (d) Explain why your answer to part (c) is not the average of the answers to parts (a) and (b).

Catching the Bus. A student is running at her top speed of 5.0 \(\mathrm{m} / \mathrm{s}\) to catch a bus, which is stopped at the bus stop. When the student is still 40.0 \(\mathrm{m}\) from the bus, it starts to pull away, moving with a constant acceleration of 0.170 \(\mathrm{m} / \mathrm{s}^{2}\) . (a) For how much time and what distance does the student have to run at 5.0 \(\mathrm{m} / \mathrm{s}\) before she overtakes the bus? (b) When she reaches the bus, how fast is the bus traveling? (c) Sketch an \(x-t\) graph for both the student and the bus. Take \(x=0\) at the initial position of the student. (d) The equations you used in part (a) to find the time have a second solution, corresponding to a later time for which the student and bus are again at the same place if they continue their specified motions. Explain the significance of this second solution. How fast is the bus traveling at this point? (e) If the student's top speed is 3.5 \(\mathrm{m} / \mathrm{s}\) , will she catch the bus? (t) What is the minimum speed the student must have to just catch up with the bus? For what time and what distance does she have to run in that case?

An object's velocity is measured to be \(v_{x}(t)=\alpha-\beta t^{2}\) where \(\alpha=4.00 \mathrm{m} / \mathrm{s}\) and \(\hat{\beta}=2.00 \mathrm{m} / \mathrm{s}^{3} .\) At \(t=0\) the object is at \(x=0\) ( a) Calculate the object's position and acceleration as functions of time. \((b)\) What is the object's maximum positive displacement from the origin?
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