Kinematics question - which equation to use?

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The discussion centers on solving a kinematics problem involving a ball thrown upward from a 50 ft building with an initial velocity of 30 ft/s. The user initially applied the equation x = (xsub0) + (vsub0)t + 1/2(a)t^2 with incorrect acceleration, leading to a discrepancy with the textbook answer. Upon using the equation \Deltax = (vsubnot)\Deltat + 1/2(a)(\Deltat)^2 with the correct acceleration of 32 ft/s², the user obtained the correct solution. The key takeaway is the importance of using consistent units and understanding when to apply each kinematic equation.

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A ball is thrown upward with velocity 30 ft/s on the roof of a 50 ft building - find the total time until the object meets the ground.

Heres my problem - I used the equation
x = (xsub0) + (vsub0)t + 1/2(a)t^2
I set xsub0 = 50, vsub0 = 30, and a = -9.8
I solved for t when x = 0, but the book gave a totally different answer.

Then I went back and used the equation given in the book:
[tex]\Delta[/tex]x = (vsubnot)[tex]\Delta[/tex]t + 1/2(a)([tex]\Delta[/tex]t)^2

I solved for [tex]\Delta[/tex]t when [tex]\Delta[/tex]x = 50, and my answer agreed
with the textbook solution...

My question is, why does the 2nd equation work in this situation, and how do I know when it is appropriate to use which?
 
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Those equations are equivalent. Used properly, you get the same answer for each. (If you got different answers, show exactly what you did.) Note that with the usual sign convention (up is +), [itex]\Delta x[/itex] should be -50.

Also: what are your units? If distance is in feet, then acceleration should be 32 ft/s^2, not 9.8 m/s^2.
 
Also: what are your units? If distance is in feet, then acceleration should be 32 ft/s^2, not 9.8 m/s^2.

i think that was my error :blushing:
thank you very much - it works out now
 

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