Posted by Alex on Friday, October 28, 2011 at 6:19pm.
An arrow is shot at 29.0° above the horizontal. Its initial speed is 54 m/s and it hits the target.
(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 7:50pm
at the max height, velocity vertical is zero.
Vf=0=54*sin29deg  g*t
solve for t. then put it in this:
hmax=54*sin29*t1/2 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 5:00pm
SIMPLIFY.
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