Answer: Kinematics Free Fall Homework Solution

AI Thread Summary
The discussion revolves around solving a kinematics problem involving a ball thrown straight up and its displacement after 6 seconds. The initial velocity was recalculated, with the correct value being 52 m/s, leading to a displacement of 132 meters. Participants clarified that the displacement should be considered zero if the ball returns to its starting point, despite the distance traveled. The equation delta y = v(t)t + 0.5gt^2 is confirmed as a valid kinematic equation that can be memorized for use with constant acceleration. The importance of maintaining a consistent coordinate system for positive and negative directions in calculations was also emphasized.
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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?
 
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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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