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 Feb 10th 2010, 07:44 AM #1 Junior Member   Join Date: Nov 2009 Posts: 9 equations of motion i have attempted a,b,c but got bit unsure about the answer for d. are my calculations correct? and if anyone could help with part d. (a) s= -u^2/2a = -25^2/2 x 9.8 = 31.9m (b) t = -u/a t=-25/-9.8 t= 2.55secs to reach max,therefore 5.1secs to return to thrower (c) v^2=u^2+2as v^2=25^2+2(-9.8x5) v=sqrt (527) v=23ms^-1 (d) this was the one i was unsure about? i tried getin an estimate from a graphical method and got 15metres @ 4secs?
 Feb 11th 2010, 09:22 PM #2 Senior Member   Join Date: Apr 2008 Location: HK Posts: 886 For d, is it simply by using s=ut + gt^2/2? s=25(4)-0.5(10)(4^2) s=20m is it the answer? __________________ Good results were achieved and the new task is to become a good doctor.

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