Kinematics question - which equation to use?

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A ball is thrown upward from a 50 ft building with an initial velocity of 30 ft/s, and the discussion centers on determining the correct kinematic equation to use for calculating the time until it hits the ground. The initial attempt used the equation x = x0 + v0t + 1/2(at^2) but yielded a different answer than the textbook. Upon switching to the equation Δx = v0Δt + 1/2(a)(Δt)^2, the user found agreement with the textbook solution. The key takeaway is the importance of consistent units, as the acceleration should be 32 ft/s² instead of 9.8 m/s² when using feet for distance. Proper application of the equations, along with correct unit conversion, leads to consistent results.
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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:
\Deltax = (vsubnot)\Deltat + 1/2(a)(\Deltat)^2

I solved for \Deltat when \Deltax = 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 +), \Delta x 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
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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