Answer: Kinematics Free Fall Homework Solution

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Homework Help Overview

The discussion revolves around a kinematics problem involving the motion of a ball thrown straight up and its subsequent displacement after a certain time. Participants are exploring the calculations related to velocity, acceleration due to gravity, and displacement.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants are attempting to calculate the displacement of the ball using kinematic equations and are questioning the signs of their results based on their coordinate system setup.
  • Some participants are clarifying whether certain equations can be used without derivation and discussing the implications of displacement in relation to the ball's motion.
  • There is a discussion about the interpretation of displacement when the ball returns to its original position.

Discussion Status

The conversation is ongoing, with participants providing insights and questioning each other's reasoning. Some guidance has been offered regarding the use of kinematic equations and the importance of consistent sign conventions in calculations.

Contextual Notes

Participants are navigating issues related to the coordinate system and the assumptions made about directionality in their calculations. There is also mention of potential confusion regarding the displacement at a specific time versus overall displacement.

madah12
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Homework Statement


. A ball is thrown straight up at t=0. It is found to be falling with a speed of 8 m/s at t= 6 sec. Find the displacement of the ball.

Homework Equations


The Attempt at a Solution


v=v0-gt
8=v0-60
v0=68
deltax =v0t - .5gt2
delta x = (68*6)- 5*36=228
but it says the answer is -132 am i wrong or is it wrong?

Edit:
nvm I got it it was supposed to be -8
-8=v0-60
52=v0
deltax=52*6 - .5*10*36 = 132

uhm it seems I messed up again this time I got a plus why isn't it a minus?
 
Last edited:
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Also I have another question is delta y = v(t)t+.5 gt^2 considered one of the kinematics equations I mean can I use it without having to derive it every time?
 
"Find the displacement of the ball."

Since you throw the ball straight up and it comes back down to where you initially threw it from, then I believe the displacement should be zero. It would be like running a lap around the track. Sure you have ran a 1/4 of a mile but your total displacement would be zero because you end up back to where you originally started.

"Also I have another question is delta y = v(t)t+.5 gt^2 considered one of the kinematics equations I mean can I use it without having to derive it every time?"

Yes that is one of the kinematic equations that you can memorize. It only applies, however, to constant accelerations.
 
i am pretty sure it meant the displacement at t=6 ...
 
madah12 said:
i am pretty sure it meant the displacement at t=6 ...

Did you use a positive acceleration or a negative acceleration? Always remember which way you set up your coordinate system (i.e. down is negative and up is positive.) If you used (-8)m/s for the velocity than we assume down is negative.
 
I used negative acceleration. - 10
 
displacement = x(final)-x(initial)
s = 0-132
When you used 52 m/s (which was your initial velocity) for your motion equation, you get an initial position.
 
x final is 132 and x initial is 0 why did you reverse it ?
 

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