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
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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