Asked by Brigid

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?

Answers

Answered by bobpursley
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
Answered by Brigid
Thanks. Thought so but my friend disagreed with me.
Answered by bobpursley
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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