[Kinematics] Calculating the maximum height reached by the ball

In summary, the conversation discusses the use of equations to solve a problem and the importance of including a minus sign in front of the acceleration. An additional tip is provided to double check the solution by calculating the position at twice the time to maximum height.
  • #1
Slimy0233
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48
Homework Statement
A ball is thrown up at a speed of 4.0 m/s. Find the
maximum height reached by the ball. Take ##g = 10 m/s^2##
Relevant Equations
##v = u +at##
##S = ut +0.5(at^2)##
I realize I can solve the other way too. But I want to solve using the equations
##v = u +at##
##S = ut +0.5(at^2)##

and I don't know why I didn't get the right answer. Thank you for your help!
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  • #2
You forgot the minus sign in front of the acceleration.
 
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  • #3
kuruman said:
You forgot the minus sign in front of the acceleration.
ahh.... my good old archnemesis: the minus sign.

Thank you for pointing that out!
 
  • #4
An additional check to this kind of problem is to calculate the position at twice the time to maximum height. It should be zero because what goes up must come down, as they say.
 
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  • #5
kuruman said:
An additional check to this kind of problem is to calculate the position at twice the time to maximum height. It should be zero because what goes up must come down, as they say.
Thank you :smile: That's helpful. I will do that from now on.
 
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