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Chapter 2: Problem 94

A rock is dropped (from rest) from the top of a \(60-\mathrm{m}\) -tall building. How far above the ground is the rock \(1.2 \mathrm{~s}\) before it reaches the ground?

Short Answer

Expert verified
The rock is 33.9 m above the ground 1.2 seconds before it impacts.

Step by step solution

01

Understand the Problem

We're asked to find the distance above the ground when a rock hits the ground at a time of \(1.2 \text{ s}\) before it actually reaches the ground. The rock falls from a height of 60 meters.
02

Determine Time of Flight

First, calculate the total time it takes for the rock to fall before it hits the ground using the formula for free fall: \( h = \frac{1}{2}gt^2 \), where \( h = 60 \text{ m} \) and \( g = 9.8 \text{ m/s}^2 \). Solve for \( t \).
03

Calculate Total Time of Flight

Set up the equation: \( 60 = \frac{1}{2} \times 9.8 \times t^2 \). This simplifies to \( 60 = 4.9t^2 \). Solving for \( t \), \( t^2 = \frac{60}{4.9} \), so \( t = \sqrt{\frac{60}{4.9}} \approx 3.51 \text{ s}\).
04

Determine Time Before Impact

Subtract \(1.2 \text{ s}\) from the total time \(3.51 \text{ s}\) to find the time of the rock's position relative to the ground. So, \( t = 3.51 - 1.2 = 2.31 \text{ s}\).
05

Calculate Distance Above Ground

Using the time \( t = 2.31 \text{ s}\), find the height above ground using the formula \( h = \frac{1}{2}gt^2 \), with \( g = 9.8 \text{ m/s}^2 \). Substitute to find \( h = \frac{1}{2} \times 9.8 \times (2.31)^2 = 26.1 \text{ m}\).
06

Compute Distance from Ground

Since the building is 60 m tall, subtract the height calculated from the total height of the building: \( 60 - 26.1 = 33.9 \text{ m}\). This is the height above the ground.

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Key Concepts

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

Gravity
When discussing free fall, gravity is the primary force at play. It pulls objects towards the center of Earth, causing them to accelerate downwards. This acceleration due to gravity is denoted by the symbol \( g \) and has a consistent value of approximately \( 9.8 \text{ m/s}^2 \) on Earth's surface.

This means that in a vacuum, where air resistance is absent, any object will increase its velocity by \( 9.8 \text{ m/s} \) every second it is falling. Because gravity鈥檚 effect is constant, it simplifies calculations when analyzing the motion of falling objects, such as the rock dropped from a building.

Understanding gravity helps explain why an object accelerates as it falls, and it is crucial in predicting how long an object will be in the air during free fall.
  • Gravity is a constant accelerating force pulling objects towards Earth.
  • The acceleration due to gravity is \( 9.8 \text{ m/s}^2 \).
Distance and Displacement
In physics, distance and displacement often describe an object's position change, but they are not the same. Distance is a scalar quantity that signifies how much ground an object has covered during its motion.

On the other hand, displacement is a vector quantity, defined as the change in position from the object's starting point to its end point. It concerns not just the distance but also the direction.

In the exercise, when calculating how far above the ground the rock is, we determine the displacement from the building's top to the rock's position at a specified time. Using the free fall equation \( h = \frac{1}{2}gt^2 \), we find how high above the ground the rock remains just before hitting the ground.
  • Distance covers the total path length, while displacement considers the initial and final points.
  • For the rock in free fall, displacement helps to determine its position above the ground at any time.
Time of Flight
The time of flight refers to the total time an object remains in the air from the moment it is released until it reaches the ground or another resting point. In free-fall scenarios, calculating this period involves solving for the time \( t \) in the equation \( h = \frac{1}{2}gt^2 \), as shown in our exercise with the falling rock.

Once we know the time it takes for the rock to hit the ground, manipulating this value provides insights into its position at different intervals before impact. By subtracting from the total time, we can determine specific moments during the rock鈥檚 descent, such as the \( 1.2 \text{ seconds} \) before it reaches the ground.

Understanding the time of flight helps predict the motion of falling objects in various conditions.
  • Time of flight measures how long an object is airborne.
  • By adjusting this time, we analyze the object's position at any point during its descent.

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Most popular questions from this chapter

To test the quality of a tennis ball, you drop it onto the floor from a height of \(4.00 \mathrm{~m}\). It rebounds to a height of \(2.00 \mathrm{~m}\). If the ball is in contact with the floor for \(12.0 \mathrm{~ms}\), (a) what is the magnitude of its average acccleration during that contact and (b) is the average acceleration up or down?

A ball is thrown vertically downward from the top of a \(36.6-\mathrm{m}\) -tall building. The ball passes the top of a window that is \(12.2 \mathrm{~m}\) above the ground \(2.00 \mathrm{~s}\) after heing thrown. What is the specd of the ball as it passes the top of the window?

A car moves along an \(x\) axis through a distance of \(900 \mathrm{~m}\), starting at rest (at \(x=0\) ) and cnding at rest (at \(x=900 \mathrm{~m}\) ). Through the first \(\frac{1}{4}\) of that distance, its acceleration is \(+2.25 \mathrm{~m} / \mathrm{s}^{2}\). Through the rest of that distance, its acceleration is \(-0.750 \mathrm{~m} / \mathrm{s}^{2}\). What are (a) its travel time through the \(900 \mathrm{~m}\) and \((\mathrm{b})\) its maximum speed? (c) Graph position \(x,\) velocity \(v,\) and acceleration \(a\) versus time \(t\) for the trip.

Suppose a rocket ship in deep space moves with con- stant acceleration equal to \(9.8 \mathrm{~m} / \mathrm{s}^{2}\), which gives the illusion of normal gravity during the flight. (a) If it starts from rest, how long will it take to acquire a speed onc-tenth that of light, which travels at \(3.0 \times 10^{8} \mathrm{~m} / \mathrm{s} ?\) (b) How far will it travel in so doing?

A car traveling \(56.0 \mathrm{~km} / \mathrm{h}\) is \(24.0 \mathrm{~m}\) from a barrier when the driver slams on the brakes. The car hits the barrier \(2.00 \mathrm{~s}\) later. (a) What is the magnitude of the car's constant acceleration before impact? (b) How fast is the car traveling at impact?
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