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natasha.aries
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***I WILL GIVE BEST ANSWER!***HEYY! CAN SOMEONE PLEASE HELP ME WITH THIS QUESTION!? I HAVE A TEST COMING UP AND I AM SOO CONFUZZLED! PLEASE INCLUDE THE NUMBERS AND SOLUTIONS IN ANSWER! (please dont round)
 9 months ago
 9 months ago
natasha.aries Group Title
***I WILL GIVE BEST ANSWER!***HEYY! CAN SOMEONE PLEASE HELP ME WITH THIS QUESTION!? I HAVE A TEST COMING UP AND I AM SOO CONFUZZLED! PLEASE INCLUDE THE NUMBERS AND SOLUTIONS IN ANSWER! (please dont round)
 9 months ago
 9 months ago

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natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
The question is what is the height of where the collision takes place!
 9 months ago

RANE Group TitleBest ResponseYou've already chosen the best response.0
r u given the answer ?
 9 months ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
sadly, no :(
 9 months ago

RANE Group TitleBest ResponseYou've already chosen the best response.0
ahan well i dis q's like this last year , i think i remember it
 9 months ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
really!? how do u do it!?
 9 months ago

RANE Group TitleBest ResponseYou've already chosen the best response.0
i think so, let me try first
 9 months ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
okay! thankk youu!
 9 months ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
i get a few tries to see if i got the question right so if you give me an answer i could put it in and see if youre right. and then you could explain it! but you cant round on anything! and this is only if youre willing to help me! sorry
 9 months ago

RANE Group TitleBest ResponseYou've already chosen the best response.0
ok i did it and i got 91.6m but im not sure if thats the answer
 9 months ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
did you round anything?
 9 months ago

RANE Group TitleBest ResponseYou've already chosen the best response.0
was i suppose to?
 9 months ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
nope! okay let me check
 9 months ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
nope its wrong :(
 9 months ago

RANE Group TitleBest ResponseYou've already chosen the best response.0
whts the answer then?
 9 months ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
idk i can put in answers and it will tell me if its right or wrong
 9 months ago

RANE Group TitleBest ResponseYou've already chosen the best response.0
idk then i cant figure out any other way to do it
 9 months ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
its okay thanks anyways!
 9 months ago

RANE Group TitleBest ResponseYou've already chosen the best response.0
np:) but dont u have nay clue wht could be the answer around ?
 9 months ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
absolutely no idea lol
 9 months ago

RANE Group TitleBest ResponseYou've already chosen the best response.0
have u done q's like that before?
 9 months ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
not as hard!
 9 months ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
this one is really complex
 9 months ago

RANE Group TitleBest ResponseYou've already chosen the best response.0
maybe he can help @UnkleRhaukus
 9 months ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
YES PLEASE @UnkleRhaukus lol
 9 months ago

oksuz_ Group TitleBest ResponseYou've already chosen the best response.0
dw:1380448186791:dw You know Vx and d. So you can calculate time t. Then you can find what the AP is by using \[\left AP \right =\frac{ g*t ^{2} }{ 2 }\] Then you can calculate BP easily. BP=ABAP
 9 months ago

CarlosGP Group TitleBest ResponseYou've already chosen the best response.1
For target:\[h_t=h_01/2g t^2\]For bullet\[h_b=v_0\sin(\theta)t1/2g t^2\]At time t_c both bodies collide:\[h_b=h_t \rightarrow v_0\sin(\theta)t_c1/2g t_c^2=h_01/2g t_c^2 \rightarrow t_c=\frac{ h_0 }{ v_0\sin(\theta) }\]and at that moment, target is at:\[h_t(t_c)=h_0\frac{ 1 }{ 2 }g \left( \frac{ h_0 }{ v_0\sin(\theta) } \right)^2=91.74.9·\left( \frac{ 91.7 }{ 109·\sin(48.4)} \right)^2\]
 9 months ago

CarlosGP Group TitleBest ResponseYou've already chosen the best response.1
and during time t_c, bullet has to cover distance "d". Let us see:\[x(t=t_c)=v_0\cos(\theta)t_c=v_0\cos(\theta)\frac{ h_0 }{ v_0\sin(\theta) }=h_0/\tan(\theta)=d\]
 9 months ago
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