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Chapter 2: Problem 39

A tennis ball on Mars, where the acceleration due to gravity is 0.379\(g\) and air resistance is negligible, is hit directly upward and returns to the same level 8.5 s later. (a) How high above its original point did the ball go? (b) How fast was it moving just after it was hit? (c) Sketch graphs of the ball's vertical position, vertical velocity, and vertical acceleration as functions of time while it's in the Martian air.

Short Answer

Expert verified
(a) 33.62 meters; (b) 15.81 m/s; (c) See described graph forms.

Step by step solution

01

Identify Given Variables and Constants

We know that the total time for the ball's upward and downward journey is 8.5 seconds. The acceleration due to gravity on Mars is given as 0.379 times Earth's gravity, which is 0.379g. Since air resistance is negligible, we can use the equations of motion. Earth's gravity, \( g \), is approximately \( 9.81 \, \text{m/s}^2 \), so Mars' gravity (\( g_M \)) is \( 0.379 \times 9.81 = 3.72 \, \text{m/s}^2 \). The ball is in the air for 4.25 seconds to reach the highest point.
02

Calculate the Maximum Height

Use the formula for vertical position: \( h = v_i t - \frac{1}{2}g_M t^2 \), where \( h \) is the height, \( v_i \) is the initial velocity, \( g_M \) is Mars' gravitational acceleration, and \( t \) is time. The ball reaches the maximum height at \( t = 4.25 \) s, where its velocity is zero, hence: \[ 0 = v_i - g_M \cdot 4.25 \rightarrow v_i = g_M \times 4.25 \]\(v_i = 3.72 \times 4.25 = 15.81 \, \text{m/s} \).The maximum height \( h \) can be found by substituting \( v_i \) back into the height equation: \[ h = 15.81 \times 4.25 - \frac{1}{2} \times 3.72 \times 4.25^2 \approx 33.62 \, \text{meters} \].
03

Calculate Initial Velocity

The maximum height calculation from Step 2 already provided the initial velocity via integration. The initial velocity of the ball is \( v_i = 15.81 \, \text{m/s} \).
04

Graphical Representation

(c) For the graph representations:- **Vertical Position vs. Time**: This graph is a parabolic curve opening downwards. It starts at the origin, rises to peak height at \( t = 4.25 \) seconds, then returns to zero at \( t = 8.5 \) seconds.- **Vertical Velocity vs. Time**: This graph is a straight line with a negative slope (down to zero), starting at \( 15.81 \, \text{m/s} \), crossing zero at \( t = 4.25 \), and ending at \( -15.81 \, \text{m/s} \) at \( t = 8.5 \) seconds.- **Vertical Acceleration vs. Time**: This graph is a constant line across the entire time interval at \( -3.72 \, \text{m/s}^2 \), representing constant acceleration due to Martian gravity.

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Key Concepts

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

Equations of Motion
The equations of motion are powerful tools in kinematics that allow us to analyze and predict the movement of objects. These equations relate key variables such as displacement, initial velocity, final velocity, acceleration, and time. Particularly, when air resistance is negligible, these equations become quite straightforward.

For vertical motion, like the tennis ball mentioned, we often use the following equations:
  • \( v = u + at \)
  • \( s = ut + \frac{1}{2}at^2 \)
  • \( v^2 = u^2 + 2as \)
Here, \( v \) is the final velocity, \( u \) is the initial velocity, \( a \) is the acceleration (gravity in free fall), \( s \) is the displacement (height in this scenario), and \( t \) is the time. The equations help in computing the unknown variables when others are given.

In our case, by setting the final velocity at the peak (where it becomes zero) in the first equation, and calculating the height with the second, we're able to thoroughly understand the object's motion on Mars.
Free Fall Motion
Free fall motion describes the movement of objects under the sole influence of gravity, without air resistance affecting them. In such cases, the only force acting on the object is gravity, which gives it a constant acceleration. This uniform acceleration simplifies calculations of the object's motion.

For the tennis ball on Mars, since we consider air resistance negligible, its motion is a classic example of free fall. The key aspects include:
  • Constant downward acceleration due to Mars's gravity.
  • Symmetry in the path: time to ascend equals time to descend.
  • Reaching zero velocity at the peak height.
This allows us to utilize the equations of motion efficiently to predict the ball's behavior both when ascending and descending. By understanding free fall, we can accurately calculate maximum height, time in the air, and the initial velocity needed to reach that height.
Gravitational Acceleration on Mars
Gravity is a force that pulls objects towards a planetary body, causing them to accelerate. On Mars, gravitational acceleration is much less than on Earth, being about 0.379 times that on Earth. This translates to a Mars gravity of approximately 3.72 m/s虏.

In our tennis ball problem, this lower gravity means that the ball will travel higher for a given initial speed than it would on Earth. It's important to understand this variation in gravitational pull when calculating motion on different planets as it directly influences the motion equations.

The implications include:
  • Longer hang time for projectiles.
  • Higher peak heights compared to a similar throw on Earth.
  • Greater vertical range of motion.
Knowing the specific acceleration due to gravity on Mars helps us in accurately predicting and analyzing the motion and trajectory of objects in its environment. It teaches us a lot about how gravity can alter kinetics across different celestial bodies.

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Most popular questions from this chapter

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof 1.00 s later. Ignore air resistance. (a) If the height of the building is 20.0 m, what must the initial speed of the first ball be if both are to hit the ground at the same time? On the same graph, sketch the positions of both balls as a function of time, measured from when the first ball is thrown. Consider the same situation, but now let the initial speed \(v_0\) of the first ball be given and treat the height \(h\) of the building as an unknown. (b) What must the height of the building be for both balls to reach the ground at the same time if (i) \(v_0\) is 6.0 m/s and (ii) \(v_0\) is 9.5 m/s? (c) If \(v_0\) is greater than some value \(v_{max}\), no value of h exists that allows both balls to hit the ground at the same time. Solve for \(v_{max}\). The value \(v_{max}\) has a simple physical interpretation. What is it? (d) If \(v_0\) is less than some value \(v_{min}\), no value of h exists that allows both balls to hit the ground at the same time. Solve for \(v_{min}\). The value \(v_{min}\) also has a simple physical interpretation. What is it?

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The acceleration of a particle is given by \(a_x(t) =\) -2.00 m/s\(^2\) + 13.00 m/s\(^3)t\). (a) Find the initial velocity \(v_{0x}\) such that the particle will have the same \(x\)-coordinate at \(t =\) 4.00 s as it had at \(t =\) 0. (b) What will be the velocity at \(t =\) 4.00 s?
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