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natasha.aries
 2 years ago
***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)
natasha.aries
 2 years ago
***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)

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natasha.aries
 2 years ago
Best ResponseYou've already chosen the best response.0The question is what is the height of where the collision takes place!

RANE
 2 years ago
Best ResponseYou've already chosen the best response.0ahan well i dis q's like this last year , i think i remember it

natasha.aries
 2 years ago
Best ResponseYou've already chosen the best response.0really!? how do u do it!?

RANE
 2 years ago
Best ResponseYou've already chosen the best response.0i think so, let me try first

natasha.aries
 2 years ago
Best ResponseYou've already chosen the best response.0okay! thankk youu!

natasha.aries
 2 years ago
Best ResponseYou've already chosen the best response.0i 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

RANE
 2 years ago
Best ResponseYou've already chosen the best response.0ok i did it and i got 91.6m but im not sure if thats the answer

natasha.aries
 2 years ago
Best ResponseYou've already chosen the best response.0did you round anything?

natasha.aries
 2 years ago
Best ResponseYou've already chosen the best response.0nope! okay let me check

natasha.aries
 2 years ago
Best ResponseYou've already chosen the best response.0nope its wrong :(

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

RANE
 2 years ago
Best ResponseYou've already chosen the best response.0idk then i cant figure out any other way to do it

natasha.aries
 2 years ago
Best ResponseYou've already chosen the best response.0its okay thanks anyways!

RANE
 2 years ago
Best ResponseYou've already chosen the best response.0np:) but dont u have nay clue wht could be the answer around ?

natasha.aries
 2 years ago
Best ResponseYou've already chosen the best response.0absolutely no idea lol

RANE
 2 years ago
Best ResponseYou've already chosen the best response.0have u done q's like that before?

natasha.aries
 2 years ago
Best ResponseYou've already chosen the best response.0this one is really complex

RANE
 2 years ago
Best ResponseYou've already chosen the best response.0maybe he can help @UnkleRhaukus

natasha.aries
 2 years ago
Best ResponseYou've already chosen the best response.0YES PLEASE @UnkleRhaukus lol

oksuz_
 2 years ago
Best ResponseYou've already chosen the best response.0dw: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

CarlosGP
 2 years ago
Best ResponseYou've already chosen the best response.1For 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:\[h_t(t_c)=h_0\frac{ 1 }{ 2 }g \left( \frac{ h_0 }{ v_0\sin(\theta) } \right)^2=91.74.9·\left( \frac{ 91.7 }{ 109·\sin(48.4)} \right)^2\]

CarlosGP
 2 years ago
Best ResponseYou've already chosen the best response.1and 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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