Answer: Kinematics Free Fall Homework Solution

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SUMMARY

The discussion focuses on solving a kinematics problem involving a ball thrown straight up and its displacement after 6 seconds. The initial velocity was calculated as 68 m/s, but the correct value was later determined to be 52 m/s after considering the downward velocity of -8 m/s. The final displacement was calculated to be -132 meters, indicating the ball fell below its starting point. The conversation also clarifies the use of the kinematic equation for displacement, confirming its validity for constant acceleration scenarios.

PREREQUISITES
  • Understanding of kinematic equations, specifically displacement and velocity formulas.
  • Knowledge of constant acceleration concepts in physics.
  • Familiarity with coordinate systems and their implications on motion analysis.
  • Ability to perform basic algebraic manipulations to solve for unknowns.
NEXT STEPS
  • Study the derivation and application of the kinematic equation: delta y = v(t)t + 0.5gt².
  • Learn about the implications of positive and negative acceleration in motion problems.
  • Explore examples of projectile motion to reinforce understanding of displacement and velocity.
  • Review coordinate system setups and their effects on motion analysis in physics.
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Students studying physics, educators teaching kinematics, and anyone looking to improve their problem-solving skills in motion-related scenarios.

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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