Posted by **Brigid** on Thursday, April 16, 2009 at 6:29pm.

If a .3kg arrow is fired at 120m/s at a 30degree angle with the horizontal, how high is it at the top of its arc? I'm confused. Do I break up the x and y components of velocity or do I set initial KE + PE = KE (top of arc)+ PE (top of arc) to get 1/2(.3)(120)*2+ 0 = 0 +.3(9.8)h and solve for h?

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**bobpursley**, Thursday, April 16, 2009 at 6:46pm
You need to break it into components.

viy=120sin30

vy at top=0=viy-gt so solve for time t to the top.

Htop= viy*t-1/2 g t^2

- Physics -
**Brigid**, Thursday, April 16, 2009 at 6:48pm
Thanks. Thought so but my friend disagreed with me.

- Physics -
**bobpursley**, Thursday, April 16, 2009 at 6:50pm
Let your friend disprove me. It is possible to use energy, but you have to find the x component of velocity first anyway.

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