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Chapter 2: Problem 71

Landing on the Moon, a spacecraft fires its rockets and comes to a complete stop just \(12 \mathrm{m}\) above the lunar surface. It then drops freely to the surface. How long does it take to fall, and what's its impact speed? (Hint: Consult Appendix E.)

Short Answer

Expert verified
The spacecraft takes approximately 4.9 seconds to hit the surface with an impact speed of about 7.84 m/s.

Step by step solution

01

Identify Known Values

The spacecraft comes to a complete stop \(12 \mathrm{m}\) above the lunar surface before dropping freely to the surface. Thus, the initial velocity \(u = 0 \mathrm{m/s}\), the distance \(s = 12 \mathrm{m}\), and the acceleration \(a\) is the acceleration due to gravity on the moon, which is approximately \(1.6 \mathrm{m/s^2}\).
02

Solve for Time Using Kinematic Equations

Using the second equation of motion: \(s = ut + \frac{1}{2} at^2\), since \(u = 0\), it simplifies to \(s = 0.5at^2\). We can rearrange the equation to find \(t = \sqrt{(2s)/a}\). Substituting our known values: \(t = \sqrt{(2 * 12m)/ 1.6 \mathrm{m/s^2}}\).
03

Compute Impact Speed Using the First Equation of Motion

Knowing \(t\), compute for the final velocity \(v\) using the first equation of motion \(v = u + at\). The initial velocity \(u\) is zero, so the equation simplifies to \(v = at\). Substitute the known value of \(a\) and the computed value of \(t\) into this equation to find the value of \(v\).

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Key Concepts

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

Free Fall Motion
When we talk about free fall motion, we're discussing a scenario where an object is moving under the influence of gravity alone, without any other forces acting on it, like air resistance or thrust. Think of it as a pure expression of gravity's pull on an object.

In the case of the lunar lander, once the rockets stop firing, it experiences free fall motion all the way down to the Moon's surface. During this descent, the only force acting on the spacecraft is the Moon's gravitational pull. Remember though, free fall doesn't mean the object is falling down at constant speed; it's actually accelerating all the time it's falling, due to the force of gravity.
Acceleration Due to Gravity
The acceleration due to gravity, often denoted by the symbol 'g', is the acceleration that Earth鈥攐r any other astronomical body鈥攊mparts on objects due to its gravitational pull. This value's commonplace here on Earth is approximately 9.81 m/s虏, but it differs depending on the celestial body you're on.

In our exercise, the context is the Moon, where gravity is much weaker, approximately 1.6 m/s虏. This significantly affects how fast objects accelerate when they're in free fall. It's crucial to use the correct value for 'g' because it determines not only how long it takes for an object to fall a certain distance but also its impact speed when it lands.
Equations of Motion
Understanding equations of motion is key to predicting how an object will move under certain forces. These equations relate an object's velocity, acceleration, the time it's been moving, and the distance it's traveled. For free-falling objects, we're often concerned with two main calculations:

  • How long it will take an object to fall a certain distance (time of descent)
  • How fast it will be moving when it hits the ground (final velocity)
These require the correct use of kinematic equations and account for initial conditions like the starting velocity and position.

For instance, in our lunar example, we had to solve for the time it took the lander to fall 12 meters. We used the equation \(s = 0.5at^2\), which directly relates distance traveled to the acceleration and time squared. The simplicity of this particular scenario鈥攚here initial velocity was zero鈥攁llowed us to use this equation neatly. The equations of motion are a set of four equations, but choosing the correct one depends on the knowns and unknowns in the situation you're analyzing.

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Most popular questions from this chapter

Consider two possible definitions of average speed: (a) the average of the values of the instantaneous speed over a time interval and (b) the magnitude of the average velocity. Are these definitions equivalent? Give two examples to demonstrate your conclusion.

You toss a book into your dorm room, just clearing a windowsill \(4.2 \mathrm{m}\) above the ground. (a) If the book leaves your hand \(1.5 \mathrm{m}\) above the ground, how fast must it be going to clear the sill? (b) How long after it leaves your hand will it hit the floor, \(0.87 \mathrm{m}\) below the windowsill?

In a drag race, the position of a car as a function of time is given by \(x=b t^{2},\) with \(b=2.000 \mathrm{m} / \mathrm{s}^{2} .\) In an attempt to determine the car's velocity midway down a 400 -m track, two observers stand at the \(180-\mathrm{m}\) and \(220-\mathrm{m}\) marks and note when the car passes. (a) What value do the two observers compute for the car's velocity over this 40 -m stretch? Give your answer to four significant figures. (b) By what percentage does this observed value differ from the instantaneous value at \(x=200 \mathrm{m} ?\)

An object's position as a function of time \(t\) is given by \(x=b t^{4}\) with \(b\) a constant. Find an expression for the instantaneous velocity, and show that the average velocity over the interval from \(t=0\) to any time \(t\) is one- fourth of the instantaneous velocity at \(t\)

In \(2009,\) Usain Bolt of Jamaica set a world record in the \(100-\mathrm{m}\) dash with a time of 9.58 s. What was his average speed?
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