Recent content by MrGoATi

yeah it kinda worked, I came real close but one additional - is there and i can't find a mistake.. so with first equation i make it sinycosy=(1+cos2x)/2 with the second. cos2x=2(sin(π/4)cosy-sinycos(π/4))2 both sin(π/4) and sin(π/4)=root2/2 and when squared = ½ cos2x=2(½sin2+½cos2-sincos)...

it's cos2x= -sin2(y) right? It does not help much though.

Forgot to say the radiant is between [0;90] oh, yeah. x2 + 7x/12 - 1 = 0 got x1=4/3 x2= - 3/4 my guess since it's tg and [0;90] is positive. tg = 4/3 and not -3/4 right? i know i can do sin = tgcos; so tgcos+cos =>(7/3)cos used tg2 = 1/cos2 cos = 3/5 and got the correct answer (7/3)*(3/5) =...

Mentor note: Moved thread to homework section ok So I'm doing supposedly easy trigonometry problems. i did the easiest ones. now I have no idea how to solve 2. first one is count sin+cos When tg - (1/tg) = -(7/12) what i figured is that i probably need to use...

ok I guess it's just too complicated for me to understand, the change of up and down makes sense, but pytagoras WT*

but g is usually used as a vector and it turns out negative most of the time does it not? did not think of this that way until now :wideeyed: in ##F= mg## ##E=mgh## ##s=at^2/2## it's not a vector so always positive? and in ##g=v-u/t## it depends on vector direction? you just made physics more...

how could getting right answer by altering values,formulas and known data be helpful? like yeah I can easily get correct answer by changing it to your suggested ideas but when I'm at exam and I do not think of that when trying to do exercise I will still get wrong answer or if I use your idea in...

sign of h is negative since down is always negative, and the travel was from higher altitude. g is usually negative but in this case it's somehow positive I got this question answered in another post, tho I'm not sure how I got multiple posts of this even though I made 1 XD.

i guess it's as if the person is on a cliff and the cliff suddenly disappears and he is hit at the top of his head which gives him extra 5m/s velocity. i got an answer on another post though, in ##E_p=mgh## ##g=10m/s^2## instead of ##g=-10m/s^2##

Homework Statement sportsman jumps at 5m/s velocity from 5m cliff to water. at what speed does he reach water? so given answer is 11.1m/s though in this book there are some wrong answers (I did try other equations and I got right answer but I want to know why I can't get it using this one)...

[ post: 5642023, member: 345636"]What you stated was not the exact statement of the problem as it was originally written. It was your paraphrasing of the problem statement. ,,ball dropped in air at 20m/s.,, that's the given information the question was ,maximum height, which i told when you...

i did completely state all that was given, and when you asked i answered what i needed to find. thanks for suggestion though i will keep that in mind

thanks, now i get it. explains a lot

yeah exactly but from a ton of forums I watched solutions to this they use s=v0xt+ gt2/2 but end up with s=gt2/2 finding that a ball thrown up with 20m/s will get to ground to maximum height in 2.04s, 20.4m the task is quite easy, but since it is so confusing for me that v0=0 I think I will be...

the task is to find maximum height, but I'm pretty sure it's 20.4 I am asking about whether velocity is v0 or v . and how do I know which is it in tasks I may have in future.