• Help me with some physics!
    32 replies, posted
[QUOTE=JohnnyMo1;25672740]Breaking it into two one-dimensional equations is super unnecessary. You can just use the range equation relaxin posted.[/QUOTE] -snipped- Doesn't come naturally for me. I use it really rarely. [editline]27th October 2010[/editline] Besides it's an extremely unintuitive formula. You either just remember it or look it up.
For Velocity Sin(2@)=9.81*500/v 1=9.81*500/v 9.81*500/1=v v=4905 For 14deg Sin(2*14)=9.81*d/4905 Sin(2*14)*4905/9.81=234.73 For 76deg Sin(2*76)=9.81*d/4905 Sin(2*76)*4905/9.81= 234.73 For 350m Sin(2*X)=9.81*350/4905 9.81*350/4905=0.7 Sin(2*X)=0.7 arcsin0.7=X/2 44.43=x/2 44.43/2=x x=22.215
[QUOTE=RELAXiN;25673268]For Velocity Sin(2@)=9.81*500/v 1=9.81*500/v 9.81*500/1=v v=4905 For 14deg Sin(2*14)=9.81*d/4905 Sin(2*14)*4905/9.81=234.73 For 76deg Sin(2*76)=9.81*d/4905 Sin(2*76)*4905/9.81= 234.73 For 350m Sin(2*X)=9.81*350/4905 9.81*350/4905=0.7 Sin(2*X)=0.7 arcsin0.7=X/2 44.43=x/2 44.43/2=x x=22.215[/QUOTE] v^2 = 4905 m^2 / s^2 v is about 70 m/s For 14 and 76 degrees the result distance looks correct In the last calculation you should have 44.43 = x*2 x = 44.43/2 x = 22.215 I didn't check it thoroughly but at a glance those seemed to be wrong. And for the angle, notice that x0 = 22.215 degrees is only one of the solutions. The other significant solution would be x1 = 45 + 22.215. Then you'd also have solutions like 22.215 + 360, 22.215 + 2*360 etc. but those probably don't matter in this case [editline]27th October 2010[/editline] You forgot to square the v there but it kind of balanced itself out in the distance calculations because you forgot it consistently
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