How to Calculate Height and Velocity of a Tennis Ball on Mars?

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Homework Help Overview

The discussion revolves around calculating the height and initial velocity of a tennis ball launched upward on Mars, where the acceleration due to gravity is significantly lower than on Earth. The problem involves the equations of motion under constant acceleration and considers the effects of time and position in the context of projectile motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of kinematic equations to determine the ball's height and initial velocity. Some express uncertainty about how to derive the answers provided in the original post, while others suggest focusing on the relationship between time and position. There are questions about the assumptions made regarding the final velocity at the peak of the ball's trajectory.

Discussion Status

The discussion is ongoing, with participants exploring various interpretations of the problem. Some have offered hints about using specific equations and the implications of the time of flight. There is a mix of confidence and uncertainty among participants regarding their understanding of the concepts involved.

Contextual Notes

Some participants note the importance of formatting in the forum and express their struggles with returning to physics after a hiatus, which may affect their confidence in discussing the problem.

SoulInNeed
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1. A tennis ball on Mars, where the acceleration due to gravity is .379g 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?



2. v (y)= v (0y) + (a (y) * t)
y=y (0) + v (0y)t + .5 a(y) t^2
v^2 (y)= v^2 (0y) + 2a(y) (y-y(0))
They're the equations of motion for constant acceleration




3. Well, the acceleration of gravity on Mars is 3.7 m/s^2. The answers are (a) 33.5 m, and (b) 15.8 m/s. I'm not sure how they got these answers, though. Any help would be greatly appreciated. Thanks!
 
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First of all, the default format in the homework section has ending bold marks for a reason. It's a bit unpleasant to read an entire bold paragraph.

Just think it through, you have the equation you need. The problem talks about time and position, you have a function for the position y(t).
Hint: say that your 8.5s = T then according to the problem y(0) = y(T)
 
Oddbio said:
First of all, the default format in the homework section has ending bold marks for a reason. It's a bit unpleasant to read an entire bold paragraph.

Just think it through, you have the equation you need. The problem talks about time and position, you have a function for the position y(t).
Hint: say that your 8.5s = T then according to the problem y(0) = y(T)
Sorry for the bad format, I'm still trying to get the hang of the forum.

OK, I'm trying to understand the logic behind this. From what I gathered in the textbook, to get the highest point, just assume that the final velocity is 0, and figure it out from there, but the original velocity needs to be known, which isn't in this case. Am I at least on the right track? Sorry for the dumbness, but I'm just getting back into Physics after multiple years away.
 
So use one of the equations to calculate the initial speed and use another to calculate height.
 
Well you could do it in pieces like that where the final velocity is zero.
In that case, what is the time for the ball to go up to it's highest point? (with no air resistance)
You shouldn't need any equations for that.
 
oddbio said:
well you could do it in pieces like that where the final velocity is zero.
In that case, what is the time for the ball to go up to it's highest point? (with no air resistance)
you shouldn't need any equations for that.
4.25?
 
OK, I got it. I feel like a complete idiot, man, lol. The initial position (y 0) is 0, and the final position (y) is 0. I use the second equation to find the initial velocity. I then plug it into the final equation, and make the final velocity 0, and I get both of them. Wow, how stupid of me.
 

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