Problem 47 - College Physics

Question

A brick is released with no initial speed from the roof of a building and strikes the ground in \(2.50 \mathrm{~s},\) encountering no appreciable air drag. (a) How tall, in meters, is the building? (b) How fast is the brick moving just before it reaches the ground? (c) Sketch graphs of this falling brick's acceleration, velocity, and vertical position as functions of time.

Step-by-Step Solution

Answer

(a) 30.64 m; (b) 24.53 m/s; (c) See graphs for acceleration, velocity, and position.

Step 1: Understand the Initial Conditions

The brick is released from a building with no initial speed, which means its initial velocity is 0 m/s. It is experiencing free fall under gravity without air resistance. The time taken to hit the ground is given as 2.50 s.

Step 2: Calculate the Height of the Building

Use the formula \( h = \frac{1}{2}gt^2 \), where \( g \) is the acceleration due to gravity (9.81 m/s²) and \( t \) is the time (2.50 s). Substitute these values:\[ h = \frac{1}{2} \times 9.81 \times (2.50)^2 = 30.6375 \text{ meters} \]Thus, the height of the building is approximately 30.64 meters.

Step 3: Calculate the Final Velocity Before Impact

Use the formula \( v = gt \), where \( g \) is 9.81 m/s² and \( t \) is 2.50 s. Calculate as follows:\[ v = 9.81 \times 2.50 = 24.525 \text{ m/s} \]So, the final velocity of the brick just before it hits the ground is approximately 24.53 m/s.

Step 4: Graphical Representation

(a) **Acceleration Graph**: Acceleration is constant at -9.81 m/s² (since downward is negative in this context). The graph is a horizontal line. (b) **Velocity Graph**: Starts at 0 m/s and becomes increasingly negative (downward) over time, forming a straight line with slope -9.81. (c) **Position Graph**: Starts at 0 and curves downward, forming a parabolic shape as it represents displacement over time due to constant acceleration.

Key Concepts

Acceleration Due to Gravity

In physics, when an object falls freely under the influence of gravity alone, it experiences a specific and constant acceleration. This is referred to as the "acceleration due to gravity" and commonly denoted by the symbol \( g \). On Earth, the standard value of \( g \) is approximately 9.81 m/s².

This means that an object in free fall will increase its velocity by 9.81 meters per second every second, assuming there is no air resistance. The direction of this acceleration is towards the center of the Earth, which is typically considered negative if you choose down as the negative direction in your coordinate system.

This constant acceleration is the same for all objects, regardless of their mass, and is a crucial factor in solving free fall problems.

Velocity Calculation

In free fall, calculating the velocity of an object just before it impacts the ground involves knowing the time it has been falling and the acceleration due to gravity. The key formula we use is:

\[ v = gt \]

Here, \( v \) is the final velocity, \( g \) is the acceleration due to gravity (9.81 m/s²), and \( t \) is the time the object has been falling. In our exercise, with a time of 2.50 seconds, the final velocity just before hitting ground is calculated as 24.53 m/s.

This velocity implies how fast the object was traveling at the moment of impact, and solving such problems helps in understanding motion under constant acceleration.

Kinematics

Kinematics is the branch of mechanics that deals with the motion of objects without considering the forces that cause the motion. During free fall, kinematics allows us to describe and predict the position and velocity of an object.

Crucial equations of kinematics for uniformly accelerated motion include:

• \( v = u + gt \)
• \( s = ut + \frac{1}{2}gt^2 \)
• \( v^2 = u^2 + 2gs \)

In these equations, \( u \) represents initial velocity (which is zero in our problem), \( v \) is final velocity, \( s \) is the displacement, and \( t \) is the time. By applying these equations, one can find parameters like the distance fallen and final velocity, as demonstrated with the brick.

Graphical Representation of Motion

Visualizing motion through graphs provides considerable insight into the changes in various parameters such as acceleration, velocity, and position over time. Let's discuss these in the context of the falling brick:

• **Acceleration Graph:** This graph is straightforward, reflecting the constant acceleration due to gravity. It is a horizontal line at -9.81 m/s², indicating a steady rate of velocity change per second, directed downwards.
• **Velocity Graph:** Starting at zero, this graph shows a straight line declining downward at a constant rate, capturing the linear increase in speed as it falls under gravity. This line's slope equals \( g \), showing constant acceleration.
• **Position Graph:** As time progresses, the graph shows an exponentially decreasing curve. This parabolic shape reflects how the object's displacement increases at a changing rate due to the consistent gravitational pull, highlighting the quadratic relationship between time and distance fallen.

Graphing these attributes helps students understand the dynamic nature of motion influenced by gravity in a more tangible and visual manner.