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ksaimouli
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 one year ago
 one year ago
ksaimouli Group Title
 one year ago
 one year ago

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ksaimouli Group TitleBest ResponseYou've already chosen the best response.1
in that b and c
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.1
@furnessj
 one year ago

furnessj Group TitleBest ResponseYou've already chosen the best response.0
do you have answers to check?
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.1
check bottom page
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.1
i did a y=112 initial velocity
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.1
y=vit4.9t^2
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.1
dw:1352405626465:dw
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.1
is that right
 one year ago

furnessj Group TitleBest ResponseYou've already chosen the best response.0
sorry i didn't look at the second page. First equation is simple: v = v_o +at 0 = 112 +(g)t now, you need to change feet into metres to use 9.81 for g 112 / 3.28 = 34.146 so 34.146 / 9.81 = t
 one year ago

furnessj Group TitleBest ResponseYou've already chosen the best response.0
Second part: you are only worried about horizontal velocity (no air resistance), the up and down plays no part here. So simply: horizontal distance = horizontal speed x time hd = 224cos30 x 7
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.1
instead of v=v0 cani use v=v0t+1/2gt^2
 one year ago

furnessj Group TitleBest ResponseYou've already chosen the best response.0
no, that equation is s = v0t + 1/2gt^2, and s is displacement, which we don't know yet
 one year ago
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