Apr 3rd 2017, 05:09 AM
Join Date: Aug 2010
First, you should explain why you did that calculation! What formula were you using and how does it help solve this problem?
What I would do:
The distance traveled with constant negative acceleration g and initial speed v is d= -(g/2)t^2+ vt. We can "complete the square" by writing this as d= -(g/2)(t^2- (2v/g)t)= -(g/2)(t^2- (2v/g)t+ v^2/g^2- v^2/g^2)= -(g/2)(t- v/g)^2+ v^2/2g. Since a square is never negative, -(g/2)(t- v/g)^2 is never positive. The maximum height occurs when t- v/g= 0 and is v^2/2g.
Solve v^2/2g= 0.254 and v^2/2g= 0.381.