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Chapter 2: Problem 42

A brick is dropped (zero initial speed) from the roof of a building. The brick strikes the ground in 1.90 s. You may ignore air resistance, so the brick is in free fall. (a) How tall, in meters, is the building? (b) What is the magnitude of the brick's velocity just before it reaches the ground? (c) Sketch \(a_y-t, v_y-t\), and \(y-t\) graphs for the motion of the brick.

Short Answer

Expert verified
The building is approximately 17.71 meters tall, and the brick's velocity just before impact is about 18.64 m/s. The graphs show constant acceleration, linear velocity increase, and a quadratic displacement curve.

Step by step solution

01

Understand the Problem

We are given a situation where a brick is dropped from a building. The initial speed is zero, and it falls freely for a time of 1.90 seconds. We need to find the height of the building, the brick's velocity before it hits the ground, and sketch the acceleration, velocity, and displacement graphs as a function of time.
02

List Known Variables

We know: - Initial velocity (\(v_i\)) = 0 m/s- Time (\(t\)) = 1.90 s- Acceleration due to gravity (\(g\)) = 9.81 m/s虏We need to find:a) Height (\(h\)) of the building in meters.b) Final velocity (\(v_f\)) just before hitting the ground.
03

Calculate the Height of the Building

Use the formula for displacement in free fall: \[ h = v_i t + \frac{1}{2} g t^2\]Substituting the given values:\[ h = 0 + \frac{1}{2} \times 9.81 \times (1.90)^2\]\[ h = 0.5 \times 9.81 \times 3.61\]\[ h \approx 17.71 \text{ meters}\]Thus, the building is approximately 17.71 meters tall.
04

Calculate Final Velocity Just Before Impact

Use the formula for final velocity in free fall:\[ v_f = v_i + g t\]Substituting the known values:\[ v_f = 0 + 9.81 \times 1.90\]\[ v_f \approx 18.639 \,\text{m/s}\]Therefore, the velocity of the brick just before it hits the ground is approximately 18.64 m/s.
05

Sketch the Graphs

To sketch the graphs:- **Acceleration vs Time (\(a_y-t\))**: This is a horizontal line above x-axis at \(a_y = 9.81 \,\text{m/s}^2\) indicating constant gravitational acceleration.- **Velocity vs Time (\(v_y-t\))**: Starts from zero and is a straight line with positive slope, reaching approximately 18.64 m/s at 1.90 seconds.- **Displacement vs Time (\(y-t\))**: A parabolic curve starting from origin, curving upward, and ending around 17.71 meters at 1.90 seconds.

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Key Concepts

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

Acceleration Due to Gravity
In free fall motion, gravity is the only force acting on an object. This creates an acceleration known as "acceleration due to gravity." At the earth's surface, it is approximately 9.81 meters per second squared. This means that for every second an object is in free fall, its velocity increases by 9.81 m/s.

This acceleration is constant and does not depend on the mass of the falling object, assuming air resistance is negligible. Because gravity is a vector quantity, it has direction as well as magnitude. In free fall scenarios, we consider gravity to be acting downwards.
  • The constant acceleration allows us to make predictions about the motion of objects in free fall using simple equations.
  • This includes calculating how fast an object will be moving after a certain period of time, and the total displacement during the fall.
Velocity-Time Graph
A velocity-time graph depicts how the velocity of an object changes over time. For a brick in free fall, this graph will be a straight line with a positive slope.

The initial velocity ( v_i ) is zero because the brick is initially at rest. As time progresses, the velocity increases linearly due to the constant acceleration of gravity ( g = 9.81 ext { m/s}^2 ). This means that the line on the velocity-time graph starts at the origin (0,0) and rises at a steady rate until it reaches the final velocity just before impact.
  • The final velocity can be calculated with the equation: v_f = v_i + g t .
  • This shows the linear relationship between velocity and time in free fall.
Displacement-Time Graph
A displacement-time graph shows how an object's position changes with time. For an object in free fall, like our brick, the curve is parabolic due to the constant acceleration of gravity.

When the initial velocity is zero, the displacement starts from zero and forms part of a parabola as the object falls. The displacement at any time can be calculated with the equation:

Displacement Formula

\( y = v_i t + \frac{1}{2} g t^2 \)
  • At t=0 , displacement y is zero because the brick has not started falling yet.
  • As time goes on, displacement increases because the brick is accelerating due to gravity.
Kinematic Equations
Kinematic equations are essential tools for solving motion problems, especially in free fall scenarios. They relate the four key variables in kinematics: displacement, initial velocity, final velocity, and time, with a constant acceleration.

For free fall motion, we use the following kinematic equations: \[ v_f = v_i + g t \] \[ y = v_i t + \frac{1}{2} g t^2 \]
The first equation helps find the final velocity after a given time, starting with an initial velocity. The second allows determination of the displacement in a given time frame.
  • Using these equations requires determining known and unknown variables. For example, in a free fall from rest, initial velocity is zero.
  • These equations simplify solving problems by eliminating the need to directly measure quantities, like velocity, during free fall.

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Most popular questions from this chapter

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

A car is stopped at a traffic light. It then travels along a straight road such that its distance from the light is given by \(x(t) = bt^2 - ct^3\), where \(b =\) 2.40 m/s\(^2\) and \(c =\) 0.120 m/s\(^3\). (a) Calculate the average velocity of the car for the time interval \(t =\) 0 to \(t =\) 10.0 s. (b) Calculate the instantaneous velocity of the car at \(t =\) 0, \(t =\) 5.0 s, and \(t =\) 10.0 s. (c) How long after starting from rest is the car again at rest?

Cars \(A\) and \(B\) travel in a straight line. The distance of \(A\) from the starting point is given as a function of time by \(x_A(t) = \alpha{t} + \beta{t}^2\), with \(\alpha =\) 2.60 m/s and \(\beta =\) 1.20 m/s\(^2\). The distance of \(B\) from the starting point is \(x_B(t) = \gamma{t}^2 - \delta{t}^3\), with \(\gamma =\) 2.80 m/s\(^2\) and \(\delta =\) 0.20 m/s\(^3\). (a) Which car is ahead just after the two cars leave the starting point? (b) At what time(s) are the cars at the same point? (c) At what time(s) is the distance from \(A\) to \(B\) neither increasing nor decreasing? (d) At what time(s) do \(A\) and \(B\) have the same acceleration?

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof 1.00 s later. Ignore air resistance. (a) If the height of the building is 20.0 m, what must the initial speed of the first ball be if both are to hit the ground at the same time? On the same graph, sketch the positions of both balls as a function of time, measured from when the first ball is thrown. Consider the same situation, but now let the initial speed \(v_0\) of the first ball be given and treat the height \(h\) of the building as an unknown. (b) What must the height of the building be for both balls to reach the ground at the same time if (i) \(v_0\) is 6.0 m/s and (ii) \(v_0\) is 9.5 m/s? (c) If \(v_0\) is greater than some value \(v_{max}\), no value of h exists that allows both balls to hit the ground at the same time. Solve for \(v_{max}\). The value \(v_{max}\) has a simple physical interpretation. What is it? (d) If \(v_0\) is less than some value \(v_{min}\), no value of h exists that allows both balls to hit the ground at the same time. Solve for \(v_{min}\). The value \(v_{min}\) also has a simple physical interpretation. What is it?

A physics teacher performing an outdoor demonstration suddenly falls from rest off a high cliff and simultaneously shouts "Help." When she has fallen for 3.0 s, she hears the echo of her shout from the valley floor below. The speed of sound is 340 m/s. (a) How tall is the cliff? (b) If we ignore air resistance, how fast will she be moving just before she hits the ground? (Her actual speed will be less than this, due to air resistance.)
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