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A ball was thrown at 50miles per second and it was to reach 3000miles what is the angle that it needs to be thrown to hit the destination? with the formula r=(1/32)(Vo)^2sin2
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hi nikko maybe theres a letter x or y next to sin2,,,like 2sin2x? so that you can solve for the double angle x.....? is it maybe like this: r=(1/32)(Vo)^2sin2x..... now you can use the logarithm process....
32r=(Vo)^2sin2x log(32r)=(2sin2x) log(Vo) sin2x=(log32r)/(2log(Vo)) 2x=arc sin((log32r)/2logVo)...now you can plug in the given datas into this formula 2x=arc sin(log(32*3000)/2log50) 2x=arc sin(4.9822/3.3974) 2x=arc sin(1.466) hmmmmm sin 1.466 is not right..its not suppose to be greater than 1
seems like something are missing in the prpblem
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