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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 being 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) Max height: 33.41 m, (b) Initial velocity: 15.78 m/s, (c) Graphs: Parabolic position, linear velocity, constant acceleration.

Step by step solution

01

Identify the known values

The acceleration due to gravity on Mars is given as \(0.379g\), where \(g = 9.8 \ m/s^2\). Therefore, the gravitational acceleration \(a = 0.379 \times 9.8 \approx 3.7142 \ m/s^2\). The total time the ball is in motion (up and down) is 8.5 seconds. We can divide this time by 2 to find the time to reach the maximum height, which is \(4.25\) seconds.
02

Calculate maximum height

Use the kinematic equation for vertical motion: \[ y = v_0 t - \frac{1}{2} at^2 \] At the peak, the final velocity \(v\) is 0. We rearrange the formula to solve for maximum height \(y\) with \(v_0 = a \cdot t\). Substitute \(a = 3.7142 \ m/s^2\) and \(t = 4.25\ \text{s}\) into the second kinematic equation: \[y_{max} = \frac{1}{2}(3.7142 \times (4.25)^2) = 33.4148 \ m\]
03

Calculate initial velocity

We can use the equation \(v_0 = a \cdot t\) to determine the initial velocity. Substitute \(a = 3.7142 \ m/s^2\) and \(t = 4.25\ \text{s}\) into \(v\): \[ v_0 = 3.7142 \times 4.25 = 15.780 \ m/s \]
04

Sketch the Graphs (Conceptual Explanation)

1. **Position vs. Time:** Start at position 0, rises to peak height at 4.25 s, and returns to 0 by 8.5 s, forming a symmetric parabolic shape. 2. **Velocity vs. Time:** Start from \(15.780 \ m/s\) decreasing to 0 \( m/s\) at 4.25 seconds, then reverses in direction to \(-15.780 \ m/s\) at 8.5 seconds, appearing as a straight line decreasing and then increasing.3. **Acceleration vs. Time:** Constant at \(-3.7142 \ m/s^2\) throughout, as gravity on Mars is constant.

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Key Concepts

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

Gravitational Acceleration
Gravitational acceleration is a key concept in physics and describes how fast an object accelerates when falling freely under the influence of gravity. On Earth, gravitational acceleration is approximately 9.8 m/s², but this value changes on other celestial bodies, such as Mars. On Mars, gravitational acceleration is much weaker at around 3.7142 m/s² (calculated from 0.379 of Earth's gravity).
This weaker gravity means that objects fall more slowly on Mars than on Earth. For students working with kinematic equations, it's important to substitute the correct gravitational acceleration value depending on the location. This means understanding the influence of gravitational force on objects in motion, especially when performing calculations for vertical motion, like the trajectory of a tennis ball hit upwards. Always consider the environment when solving physics problems, since gravitational acceleration greatly determines the motion outcomes.
Vertical Motion
Vertical motion in physics refers to the movement of an object in a straight line up or down. This type of motion is particularly straightforward when there's no air resistance to consider, making the equations of motion simpler. The key equations used here are the kinematic equations, which help calculate various parameters like velocity, height, and time.
In the case of the tennis ball on Mars, vertical motion involves the ball being hit upwards, slowing to a stop at its peak, and then descending back to its original height. The symmetric nature of this vertical path is crucial because it allows us to divide the total time of motion evenly to find out how long it takes to reach the peak height.
  • Position vs. Time Graph: This curve shows the ascent and descent of the ball in a parabolic motion.
  • Velocity vs. Time Graph: Demonstrates a linear decrease in velocity to zero at the peak, then an increase in negative velocity.
Understanding vertical motion helps in comprehending how objects behave under the influence of gravity alone, without other forces like air resistance altering the path.
Initial Velocity
Initial velocity is the speed at which an object begins its motion. Calculating the initial velocity is a pivotal step in determining how high an object will go when moving vertically. In our scenario, we found the initial velocity of the tennis ball by using the simple kinematic equation: \( v_0 = a \cdot t \).
Here, \( a \) is the gravitational acceleration on Mars, and \( t \) is the time taken to reach the maximum height. This step is crucial as it directly impacts how far and how long the object will travel upwards before gravity pulls it back down. Having a clear understanding of initial velocity allows for accurate predictions of maximum heights and times of flight.
In practical terms, if you know how long something is moving up or down or under what acceleration, you can often determine the initial speed needed for such a motion. Remember, though, that this is under ideal conditions, such as negligible air resistance, which is a notable condition in this problem.
Maximum Height
The maximum height is the highest point an object reaches when it is thrown upwards. It's a critical part of understanding vertical motion because it quantifies how far an object can go against the force of gravity. To find the maximum height reached by our tennis ball on Mars, we use the kinematic equation:
\[ y = v_0 t - \frac{1}{2} at^2 \]In this formula, \( y \) represents the maximum height, \( v_0 \) is the initial velocity, \( t \) is the time to reach the maximum height, and \( a \) is the acceleration due to gravity. For our tennis ball, this equation tells us that the ball shot up to approximately 33.4148 meters.
This calculation not only illustrates how the initial speed impacts the trajectory and height but also shows how gravity works as a constant decelerating force during upward motion. Grasping how maximum height is derived can deliver insight into various real-world scenarios, such as launching projectiles or planning space missions where vertical lift is crucial.

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Most popular questions from this chapter

A 15-kg rock is dropped from rest on the earth and reaches the ground in 1.75 \(\mathrm{s}.\) When it is dropped from the same height on Saturn's satellite Enceladus, it reaches the ground in 18.6 \(\mathrm{s}.\) What is the acceleration due to gravity on Enceladus?

Determined to test the law of gravity for himself, a student walks off a skyscraper 180 \(\mathrm{m}\) high, stopwatch in hand, and starts his free fall (zero initial velocity). Five seconds later, Superman arrives at the scene and dives off the roof to save the student. Superman leaves the roof with an initial speed \(v_{0}\) that he produces by pushing himself downward from the edge of the roof with his legs of steel. He then falls with the same acceleration as any freely falling body. (a) What must the value of \(v_{0}\) be so that Superman catches the student just before they reach the ground? (b) On the same graph, sketch the positions of the student and of Superman as functions of time. Take Superman's initial speed to have the value calculated in part (a). (c) If the height of the skyscraper is less than some minimum value, even Superman can't reach the student before he hits the ground. What is this minimum height?

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