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Chapter 2: Problem 38

A brick is dropped (zero initial speed) from the roof of a building. The brick strikes the ground in \(1.90 \mathrm{~s}\). 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 height of the building is approximately \(17.2 m\), and the magnitude of the velocity of the brick just before it hits the ground is approximately \(18.6 m/s\) . The a_y-t graph is a horizontal line at -9.8 m/s^2, the v_y-t graph is a straight line starting from zero and with negative slope, and y-t graph is a downward opening parabola.

Step by step solution

01

Find the height of the building

The displacement (height of the building) can be found using the equation for displacement under constant acceleration: \(d = v_{i}t + 0.5gt^{2}\), where \(v_{i}\) is the initial velocity (zero in this case), \(g\) is the acceleration due to gravity and \(t\) is the time. \n Substitute the known values into the equation to get \(d = (0 * 1.90) + 0.5 * 9.8 * (1.90)^{2}\)
02

Calculate the final velocity

The final velocity can be calculated using the formula \(v_{f} = v_{i} + gt\). Substitutute the known values to get \(v_{f} = 0 + (9.8 m/s^{2} * 1.9 s)\)
03

Sketch the graphs

To sketch the graphs, we can use the general shapes of the displacement (y-t),velocity (v_y-t) and acceleration (a_y-t) versus time graphs for bodies under constant acceleration. The a_y-t graph is a horizontal line equal to the negative of the acceleration due to gravity, since the acceleration is constant. The v_y-t graph is a line with negative slope, starting from zero. The y-t graph is a downward opening parabola since the acceleration is constant and the object starts from rest.

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Key Concepts

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

Displacement Under Constant Acceleration
Understanding displacement in a physics context is essential for solving problems related to motion. In the case of free fall, displacement refers to the vertical distance an object travels while under the influence of gravity's constant acceleration. To determine this distance, or the height of a building from which a brick is dropped, as in our example, we use the formula:

\[ d = v_i t + \frac{1}{2}gt^2 \]

where \( d \) is the displacement, \( v_i \) is the initial velocity, \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)), and \( t \) is the time during which the object is in motion. Since our brick starts from rest, \( v_i \) equals zero, simplifying our calculation. Through this equation, we find that the acceleration acts over the time period to cover a specific distance, which is the height of the building in our problem.
Final Velocity Calculation
Knowing the final velocity of an object in free fall right before impact is vital for many applications, including safety calculations and understanding impact forces. To compute the final velocity of our brick before it hits the ground, we employ the equation:

\[ v_f = v_i + gt \]

In this formula, \( v_f \) represents the final velocity, \( v_i \) the initial velocity, \( g \) the acceleration due to gravity, and \( t \) the time elapsed. Given that the brick started its journey from rest, the initial velocity is zero, making our calculation straightforward. The final velocity is simply the product of the acceleration due to gravity and the time time falling. Mental visualization of the growing speed, or looking at a velocity-time graph, aids in grasping this concept.
Motion Graphs
Motion graphs are a powerful tool to visualize and understand the kinematics of moving objects. For the brick in free fall, three types of graphs are commonly used: acceleration-time (\(a_y-t\)), velocity-time (\(v_y-t\)), and displacement-time (\(y-t\)).

Acceleration-Time (\(a_y-t\)) Graph

For an object in free fall, the acceleration due to gravity is constant, meaning the \(a_y-t\) graph is a horizontal line that reflects the value of gravitational acceleration, in this case, -9.8 m/s^2. The negative sign indicates the acceleration is directed downwards.

Velocity-Time (\(v_y-t\)) Graph

The \(v_y-t\) graph for a free-falling object is a straight, sloped line. It begins at zero and slopes downwards, indicating a steady increase in downward speed (negative velocity) over time due to constant acceleration..

Displacement-Time (\(y-t\)) Graph

The \(y-t\) graph shows the displacement of the object over time. It forms a parabola opening downward, starting from the initial height, and curving down to zero where the brick hits the ground. The shape of the curve reflects the constant acceleration, with the steepness increasing over time as the brick's speed increases.

By sketching and interpreting these graphs, students gain a clearer, visual understanding of the kinematics involved in free fall.

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Most popular questions from this chapter

The acceleration of a particle is given by \(a_{x}(t)=\) \(-2.00 \mathrm{~m} / \mathrm{s}^{2}+\left(3.00 \mathrm{~m} / \mathrm{s}^{3}\right) t\) Find the initial velocity \(v_{0 x}\) such that the particle will have the same \(x\) coordinate at \(t=4.00 \mathrm{~s}\) as it had at t=0 . \text { (b) What will be the velocity at } t=4.00 \mathrm{~s} ?

A rock is thrown straight up with an initial speed of \(24.0 \mathrm{~m} / \mathrm{s}\) Neglect air resistance. (a) At \(t=1.0 \mathrm{~s}\), what are the directions of the velocity and acceleration of the rock? Is the speed of the rock increasing or decreasing? (b) At \(t=3.0 \mathrm{~s}\), what are the directions of the velocity and acceleration of the rock? Is the speed of the rock increasing or decreasing?

You are on the roof of the physics building, \(46.0 \mathrm{~m}\) above the ground (Fig. \(\mathbf{P 2 . 7 0}\) ). Your physics professor, who is \(1.80 \mathrm{~m}\) tall, is walking alongside the building at a constant speed of \(1.20 \mathrm{~m} / \mathrm{s}\). If you wish to drop an egg on your professor's head, where should the professor be when you release the egg? Assume that the egg is in free fall.

A car and a truck start from rest at the same instant, with the car initially at some distance behind the truck. The truck has a constant acceleration of \(2.10 \mathrm{~m} / \mathrm{s}^{2},\) and the car has an acceleration of \(3.40 \mathrm{~m} / \mathrm{s}^{2}\). The car overtakes the truck after the truck has moved \(60.0 \mathrm{~m}\). (a) How much time does it take the car to overtake the truck? (b) How far was the car behind the truck initially? (c) What is the speed of each when they are abreast? (d) On a single graph, sketch the position of each vehicle as a function of time. Take \(x=0\) at the initial location of the truck.

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. \(\mathbf{E} 2.42\) ). After the sandbag is released, it is in free fall. (a) Compute the position and velocity of the sandbag at \(0.250 \mathrm{~s}\) and \(1.00 \mathrm{~s}\) after its release. (b) How many seconds after its release does 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.
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