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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)

Physics
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The question is what is the height of where the collision takes place!
r u given the answer ?

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Other answers:

sadly, no :(
ahan well i dis q's like this last year , i think i remember it
really!? how do u do it!?
i think so, let me try first
okay! thankk youu!
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
ok i did it and i got 91.6m but im not sure if thats the answer
did you round anything?
nop
was i suppose to?
nope! okay let me check
nope its wrong :(
whts the answer then?
idk i can put in answers and it will tell me if its right or wrong
idk then i cant figure out any other way to do it
its okay thanks anyways!
np:) but dont u have nay clue wht could be the answer around ?
absolutely no idea lol
have u done q's like that before?
not as hard!
this one is really complex
sure it is
maybe he can help @UnkleRhaukus
YES PLEASE @UnkleRhaukus lol
|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|=|AB|-|AP|
For target:\[h_t=h_0-1/2g t^2\]For bullet\[h_b=v_0\sin(\theta)t-1/2g t^2\]At time t_c both bodies collide:\[h_b=h_t \rightarrow v_0\sin(\theta)t_c-1/2g t_c^2=h_0-1/2g t_c^2 \rightarrow t_c=\frac{ h_0 }{ v_0\sin(\theta) }\]and at that moment, target is at:\[h_t(t_c)=h_0-\frac{ 1 }{ 2 }g \left( \frac{ h_0 }{ v_0\sin(\theta) } \right)^2=91.7-4.9·\left( \frac{ 91.7 }{ 109·\sin(48.4)} \right)^2\]
and during time t_c, bullet has to cover distance "d". Let us see:\[x(t=t_c)=v_0\cos(\theta)t_c=v_0\cos(\theta)\frac{ h_0 }{ v_0\sin(\theta) }=h_0/\tan(\theta)=d\]

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