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Old Feb 10th 2010, 07:44 AM   #1
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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?
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Old Feb 11th 2010, 09:22 PM   #2
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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?
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