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Chapter 2: Problem 48

A hoodlum throws a stone vertically downward with an initial speed of \(12.0 \mathrm{~m} / \mathrm{s}\) from the roof of a building, \(30.0 \mathrm{~m}\) above the ground. (a) How long does it take the stone to reach the ground? (b) What is the speed of the stone at impact?

Short Answer

Expert verified
(a) The stone takes approximately 1.568 s to reach the ground. (b) The speed at impact is 27.4 m/s.

Step by step solution

01

Understand the Problem

We need to determine the time it takes for the stone to reach the ground and its speed upon impact. Given: initial velocity \(v_i = 12.0\, \mathrm{m/s}\) downward, initial height \(h = 30.0\, \mathrm{m}\), and acceleration due to gravity \(g = 9.81\, \mathrm{m/s^2}\).
02

Use the Kinematic Equation for Time

We will use the kinematic equation for vertical motion, \( h = v_it + \frac{1}{2}gt^2 \), where \(h\) is the height, \(v_i\) is the initial velocity, \(g\) is the acceleration due to gravity, and \(t\) is the time. Substitute the known values: \(30.0 = 12.0t + \frac{1}{2} \times 9.81t^2\).
03

Solve the Quadratic Equation for Time

Rearranging the equation gives \(4.905t^2 + 12.0t - 30.0 = 0\). This is a quadratic equation of the form \(at^2 + bt + c = 0\) where \(a = 4.905\), \(b = 12.0\), and \(c = -30.0\). Use the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to solve for \(t\).
04

Calculate Time Using the Quadratic Formula

Calculate the discriminant: \(b^2 - 4ac = 12.0^2 - 4 \times 4.905 \times (-30.0) = 576.9\). Now calculate \(t = \frac{-12.0 \pm \sqrt{576.9}}{2 \times 4.905}\). We take the positive root, so \(t \approx 1.568\, \mathrm{s}\).
05

Use Kinematic Equation for Final Velocity

Now, to find the final velocity, use the equation \(v_f = v_i + gt\). Substitute: \(v_f = 12.0 + 9.81 \times 1.568\).
06

Calculate Final Velocity

Substituting in the values, we get \(v_f = 12.0 + 9.81 \times 1.568 \approx 27.4 \mathrm{~m/s}\). The speed of the stone at impact is \(27.4 \mathrm{~m/s}\).

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Key Concepts

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

Kinematic Motion
Kinematic motion involves the study of the motion of objects without considering the forces that cause this motion. The core idea here is to understand how various physical quantities are interrelated when an object is in motion. For a stone thrown vertically downward, like in our exercise, a few key terms help outline this motion:

  • Initial Velocity (\(v_i\)): This is the speed at which the object begins its motion. For our example, it鈥檚 given as \(12.0\,\mathrm{m/s}\) downward.
  • Acceleration due to Gravity (\(g\)): Earth's gravity accelerates objects downward at \(9.81\,\mathrm{m/s^2}\).
  • Time (\(t\)): This is the duration the object is in motion until it reaches the ground.
  • Displacement (\(h\)): This represents the vertical distance traveled, which here is \(30.0\,\mathrm{m}\) downward.

Kinematic equations help link these quantities together, enabling us to solve for unknown variables once we know the others.
Quadratic Equations
Quadratic equations appear frequently in physics, especially in kinematics, when dealing with vertically moving objects. In our problem, we derived a quadratic equation from the vertical motion formula. This equation allows us to solve for the time, illustrating a typical method of handling kinematic problems:

  • We used the kinematic equation: \[ h = v_i t + \frac{1}{2}g t^2 \]
  • This translates to finding time \(t\) using the equation: \[ 4.905t^2 + 12.0t - 30.0 = 0 \]
  • This is in the standard quadratic form \(at^2 + bt + c = 0\), where the constants are derived from our given values.
  • To find \(t\), we relied on the quadratic formula: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
  • The discriminant \(b^2 - 4ac\) helps determine the nature of the roots, ensuring a real positive root exists for our time duration.

Solving this provides the time \(t = 1.568\,\mathrm{s}\), showing how quadratic mathematics plays a crucial role in calculating motion parameters.
Final Velocity
Final velocity in the context of kinematic motion refers to the speed an object attains just before it hits a surface. It's essential to understand how initial speeds, gravity, and time contribute to this final speed.


To determine the stone's final velocity upon impact, we use the equation:
  • \(v_f = v_i + gt\)
  • "Initial velocity" (\(v_i\)) is \(12.0\,\mathrm{m/s}\).
  • "Time" (\(t\)) calculated from our quadratic equation is \(1.568\,\mathrm{s}\).
  • "Acceleration due to gravity" (\(g\)) is \(9.81\,\mathrm{m/s^2}\).
Thus, substituting these values gives:

\[ v_f = 12.0 + 9.81 \times 1.568 \approx 27.4\,\mathrm{m/s}\]

This result signifies the speed at which the stone strikes the ground, emphasizing how initial movement and gravitational pull combine to define the object's final velocity in free fall.

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Most popular questions from this chapter

When a high-speed passenger train traveling at \(161 \mathrm{~km} / \mathrm{h}\) rounds a bend, the engineer is shocked to see that a locomotive has improperly entered onto the track from a siding and is a distance \(D=676 \mathrm{~m}\) ahead (Fig. 2-32). The locomotive is moving at \(29.0 \mathrm{~km} / \mathrm{h}\). The engineer of the high-speed train immediately applies the brakes. (a) What must be the magnitude of the resulting constant deceleration if a collision is to be just avoided? (b) Assume that the engineer is at \(x=0\) when, at \(t=0\), he first spots the locomotive. Sketch \(x(t)\) curves for the locomotive and high-speed train for the cases in which a collision is just avoided and is not quite avoided.

Two subway stops are separated by \(1100 \mathrm{~m}\). If a subway train accelerates at \(+1.2 \mathrm{~m} / \mathrm{s}^{2}\) from rest through the first half of the distance and decelerates at \(-1.2 \mathrm{~m} / \mathrm{s}^{2}\) through the second half, what are (a) its travel time and (b) its maximum speed? (c) Graph \(x, v\), and \(a\) versus \(t\) for the trip.

A shuffleboard disk is accelerated at a constant rate from rest to a speed of \(6.0 \mathrm{~m} / \mathrm{s}\) over a \(1.8 \mathrm{~m}\) distance by a player using a cue. At this point the disk loses contact with the cue and slows at a constant rate of \(2.5 \mathrm{~m} / \mathrm{s}^{2}\) until it stops. (a) How much time elapses from when the disk begins to accelerate until it stops? (b) What total distance does the disk travel?

A hot-air balloon is ascending at the rate of \(12 \mathrm{~m} / \mathrm{s}\) and is \(80 \mathrm{~m}\) above the ground when a package is dropped over the side. (a) How long does the package take to reach the ground? (b) With what speed does it hit the ground?

In \(1 \mathrm{~km}\) races, runner 1 on track 1 (with time \(2 \mathrm{~min}, 27.95 \mathrm{~s}\) ) appears to be faster than runner 2 on track \(2(2 \mathrm{~min}, 28.15 \mathrm{~s})\). However, length \(L_{2}\) of track 2 might be slightly greater than length \(L_{1}\) of track 1. How large can \(L_{2}-L_{1}\) be for us still to conclude that runner 1 is faster?
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