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A brick is dropped (zero initial speed) from the roof of a building. The brick strikes the ground in 2.50 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
(a) 30.66 m (b) 24.53 m/s (c) Sketch horizontal acceleration, linear velocity, and parabolic height graphs.

Step by step solution

01

Determine the acceleration

Since the brick is in free fall, the acceleration is due to gravity. On Earth, the value of acceleration due to gravity \( g \) is approximately \( 9.81 \, \text{m/s}^2 \). This constant acceleration \( a = 9.81 \, \text{m/s}^2 \) will be used in all subsequent calculations.
02

Calculate the height of the building

We can use the kinematic equation for displacement:\[ y = v_i t + \frac{1}{2} a t^2 \]where \( y \) is the displacement (height of the building), \( v_i = 0 \, \text{m/s} \) is the initial velocity, \( a = 9.81 \, \text{m/s}^2 \) is the acceleration, and \( t = 2.50 \, s \) is the time. Plugging in these values:\[ y = 0 + \frac{1}{2} \times 9.81 \, \text{m/s}^2 \times (2.50 \, ext{s})^2 \]\[ y = \frac{1}{2} \times 9.81 \times 6.25 \]\[ y = 30.65625 \, \text{m} \]Therefore, the height of the building is approximately 30.66 meters.
03

Calculate the velocity just before impact

To find the velocity of the brick just before it hits the ground, use the equation:\[ v_f = v_i + at \]where \( v_f \) is the final velocity, \( v_i = 0 \, \text{m/s} \), \( a = 9.81 \, \text{m/s}^2 \), and \( t = 2.50 \, s \). Plug these values into the equation:\[ v_f = 0 + 9.81 \, \text{m/s}^2 \times 2.50 \, s \]\[ v_f = 24.525 \, \text{m/s} \]Thus, the magnitude of the brick's velocity just before reaching the ground is approximately 24.53 m/s.
04

Sketch graphs of the motion

**(a) Acceleration vs Time (\( a_y - t \))**: The graph is a horizontal line at \( a_y = 9.81 \, \text{m/s}^2 \) since the acceleration is constant during free fall. **(b) Velocity vs Time (\( v_y - t \))**: The graph is a straight line starting from the origin (0,0) with a positive slope of 9.81. It ends at \( v_y = 24.53 \, \text{m/s} \) at \( t = 2.50 \, s \).**(c) Height vs Time (\( y - t \))**: The graph is a parabolic curve. It starts at \( y = 0 \) at \( t = 0 \) and reaches \( y = 30.66 \, \text{m} \) at \( t = 2.50 \, s \). The curve is downward opening due to the negative direction of displacement (falling down).

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Key Concepts

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

Free Fall
Free fall is a type of motion in which an object is only influenced by gravity. When an object is in free fall, it experiences acceleration due to gravity, with other forces such as air resistance being negligible. This means that any object freely falling under the influence of gravity will have a uniform acceleration towards the Earth.

In our example of the brick being dropped from a building, it starts from rest, with no initial velocity. Thus, it is purely under the effect of gravitational pull. The constant acceleration of the brick constitutes what we commonly refer to as free fall. It's noteworthy that free fall doesn't necessarily imply a lack of velocity—just that the only force acting on the object is gravity.

Remember, every object regardless of its mass accelerates at the same rate when in free fall. Therefore, both a feather and a brick would, theoretically in a vacuum, fall at the same time from the same height.
Acceleration Due to Gravity
In the context of kinematics, the acceleration due to gravity is denoted by the symbol \( g \). On Earth, this value is approximately \( 9.81 \, \text{m/s}^2 \). This means that for every second an object is in free fall, its velocity increases by \( 9.81 \, \text{m/s} \), assuming air resistance is negligible.

When solving physics problems involving objects in free fall, such as the dropping brick from the building, it is essential to apply \( g \) as the acceleration. In these calculations, knowing \( g \) allows us to compute both the distance the object travels and its speed at a given time while falling.

Since \( g \) is consistent, it provides a reliable basis for predictions and calculations in any free-fall scenario. The magnitude remains constant at \( 9.81 \, \text{m/s}^2 \), no matter how high the object starts from.
Kinematic Equations
Kinematic equations are crucial in solving problems related to motion, especially in free fall. They simplify the calculation of variables such as distance, time, and velocity. In the scenario involving the brick, the following kinematic equations are applied:

  • To find the height of the building, we use:
    \[ y = v_i t + \frac{1}{2} a t^2 \]
    where \( y \) is the displacement, and \( a \) is the acceleration.

  • To find the final velocity just before impact, we apply:
    \[ v_f = v_i + at \]
    where \( v_f \) is the final velocity.

These equations involve basic parameters: initial velocity \( v_i \), final velocity \( v_f \), acceleration \( a \), time \( t \), and displacement \( y \). Each problem requires substituting known values to solve for unknown variables. Mastery in using kinematic formulas allows anyone to analyze and predict the motion of freely falling objects effectively.
Velocity-Time Graph
A velocity-time graph is a powerful tool for visualizing an object's motion over time. In such a graph, velocity is plotted on the vertical axis, while time is on the horizontal axis. For a brick in free fall, the velocity-time graph presents as a straight line:

  • The graph starts at the origin point, indicating the initial velocity \( v_i \) is zero.

  • It has a positive slope of \( 9.81 \, \text{m/s}^2 \), corresponding to the constant acceleration due to gravity.

  • The line extends up to \( v_f = 24.53 \, \text{m/s} \) at \( t = 2.50 \, \text{s} \).

The consistent slope reflects the unchanging acceleration. You can determine the velocity at any point during the fall by simply looking at the graph. The more time passes, the higher the velocity increases, linearly.

This representation helps to better understand how acceleration maintains its role in increasing velocity over time during free fall.

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