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October 1, 2014

October 1, 2014

Posted by **Alex** on Friday, October 28, 2011 at 6:19pm.

(a) What is the maximum height the arrow will attain?

(b) The target is at the height from which the arrow was shot. How far away is it?

- Physics -
**bobpursley**, Friday, October 28, 2011 at 7:50pmat the max height, velocity vertical is zero.

Vf=0=54*sin29deg - g*t

solve for t. then put it in this:

hmax=54*sin29*t-1/2 g t^2

distance target:= 54cos29*(2t) where t is the time to the max height, 2t is the total time in the air.

- Physics -
**Alex**, Sunday, October 30, 2011 at 5:00pmSIMPLIFY.

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