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March 5, 2015

March 5, 2015

Posted by **Anonymous** on Saturday, November 6, 2010 at 10:51pm.

- physics -
**hy**, Sunday, November 7, 2010 at 12:11amintial Vy=0 (since it's thrown horizontal) change in y = -29.0m Ay=-9.81 m/s^2

find the time used by equate :

change in Y= intial Vy times time +1/2 Ay times time squared so -29.0=1/2(-9.81)t^2

t^2 = .91s so t=2.43s

since time used for y is same as x

change in x = 1/2 (intial Vx+ final Vx) times time

in progectile motion ini Vx = fin Vx

so change in x = time(2.43s) times intial vx (4.00m/s))

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