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?

Answer 1

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

Answer 2

Thanks. Thought so but my friend disagreed with me.

Answer 3

Let your friend disprove me. It is possible to use energy, but you have to find the x component of velocity first anyway.