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Chapter 2: Problem 50

A woman is reported to have fallen 144 ft from the 17 th floor of a building, landing on a metal ventilator box, which she crushed to a depth of 18.0 in. She suffered only minor injuries. Neglecting air resistance, calculate (a) the speed of the woman just before she collided with the ventilator, (b) her average acceleration while in contact with the box, and (c) the time it took to crush the box.

Short Answer

Expert verified
The speed of the woman just before she collided with the ventilator is approx 29.42 m/s. Her average acceleration while in contact with the box is 306.9 m/s^2. The time it took to crush the box was approx 0.096 seconds.

Step by step solution

01

Calculation of the speed of the woman before she collided with the box

Assuming she started from rest and fell under the influence of gravity alone, we can use the following formula from physics for objects in free falling: \(v = \sqrt{2gh}\). Since we need to convert feet to meters, we set \(h = 144 ft = 43.9 m\), and \(g = 9.81 m/s^2\). Plugging into the formula: \(v = \sqrt{2*9.81*43.9} = 29.42 m/s\).
02

Calculation of average acceleration while in contact with the box

The woman's final speed after hitting the box is considered as zero. According to the formula for acceleration: \(a = \Delta v/ \Delta t\). We can consider \(\Delta v = v - 0 = 29.42 m/s\). We also convert 18 in depth of the box to meters as \(\Delta x = 18 in = 0.46 m\). Then we use the second equation of motion: \(\Delta x = v*\Delta t - 0.5* a* (\Delta t)^2\). Solving this quadratic equation yields the roots corresponding to the two impacts (bottom and the bounce). The positive root gives the time \(\Delta t = 0.096 s\). Then we can find the average acceleration \(a\) = \(\Delta v/ \Delta t\) = 29.42 m/s / 0.096 s = 306.9 m/s^2.
03

Calculation of the time it took to crush the box

The time it took to crush the box is the same as the time during which the average acceleration occurred, which we found as \(\Delta t = 0.096 s\) in the previous step.

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Key Concepts

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

Kinematics
Kinematics is a branch of mechanics that deals with the motion of objects without considering the forces that cause this motion. It involves the study of positions, velocities, and accelerations of objects, and it is fundamental to understand free fall physics. To describe the motion of objects, kinematic equations are used, such as the formula for the final velocity of an object in free fall, given by
\( v = \sqrt{2gh} \),
where \( v \) is the final velocity, \( g \) is the acceleration due to gravity, and \( h \) is the height from which the object falls. These equations apply under the assumption that there is no air resistance and the only force acting on the object is gravity. This is why in the example of the woman falling from a building, the kinematic formula could be applied to calculate her speed just before impact.
Acceleration Due to Gravity
The acceleration due to gravity, denoted by \( g \), is the acceleration gained by an object due to the gravitational force of Earth at its surface. It is approximately \( 9.81 m/s^2 \) and is a constant in free fall calculations. This constant allows us to predict how an object will move when it is falling towards the Earth's surface. Understanding this acceleration is crucial in physics calculations as it helps us describe the behavior of a freely falling body through kinematic equations. In our textbook example, the woman falling from a building experiences this acceleration, ignoring air resistance, which makes her case an excellent scenario to apply the concept of free fall.
Equations of Motion
The equations of motion are a set of formulas that allow us to relate velocity, acceleration, time, and displacement of a moving object. There are three main equations, but in free fall problems, we often use the formula that relates final velocity, acceleration, and displacement:
\( v^2 = u^2 + 2as \),
where \( v \) is the final velocity, \( u \) is the initial velocity (which is often zero in free fall cases), \( a \) is the acceleration, and \( s \) is the displacement. When calculating the average acceleration of the woman after hitting the ventilator box, another motion equation is utilized:
\( s = ut + \frac{1}{2}at^2 \),
which helps to find the time and acceleration during the period she crushes the box. These formulas are vital tools for physicists and engineers to predict the future position and velocity of objects in motion.

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Most popular questions from this chapter

An inquisitive physics student and mountain climber climbs a 50.0 -m cliff that overhangs a calm pool of water. He throws two stones vertically downward, \(1.00 \mathrm{s}\) apart, and observes that they cause a single splash. The first stone has an initial speed of \(2.00 \mathrm{m} / \mathrm{s} .\) (a) How long after release of the first stone do the two stones hit the water? (b) What initial velocity must the second stone have if they are to hit simultaneously? (c) What is the speed of each stone at the instant the two hit the water?

The speed of a bullet as it travels down the barrel of a rifle toward the opening is given by \(v=\left(-5.00 \times 10^{7}\right) t^{2}+\) \(\left(3.00 \times 10^{5}\right) t,\) where \(v\) is in meters per second and \(t\) is in seconds. The acceleration of the bullet just as it leaves the barrel is zero. (a) Determine the acceleration and position of the bullet as a function of time when the bullet is in the barrel. (b) Determine the length of time the bullet is accelerated. (c) Find the speed at which the bullet leaves the barrel. (d) What is the length of the barrel?

Draw motion diagrams for (a) an object moving to the right at constant speed, (b) an object moving to the right and speeding up at a constant rate, \((c)\) an object moving to the right and slowing down at a constant rate, (d) an object moving to the left and speeding up at a constant rate, and (e) an object moving to the left and slowing down at a constant rate. (f) How would your drawings change if the changes in speed were not uniform; that is, if the speed were not changing at a constant rate?

A freely falling object requires \(1.50 \mathrm{s}\) to travel the last \(30.0 \mathrm{m}\) before it hits the ground. From what height above the ground did it fall?

Setting a new world record in a \(100-\mathrm{m}\) race, Maggie and Judy cross the finish line in a dead heat, both taking \(10.2 \mathrm{s}\) Accelerating uniformly, Maggie took \(2.00 \mathrm{s}\) and \(\mathrm{Judy} 3.00 \mathrm{s}\) to attain maximum speed, which they maintained for the rest of the race. (a) What was the acceleration of each sprinter? (b) What were their respective maximum speeds? (c) Which sprinter was ahead at the \(6.00-\) s mark, and by how much?
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