A community for students.
Here's the question you clicked on:
 0 viewing
IsTim
 4 years ago
A medieval prince trapped in a castle wraps a message around a rock and throws it from the top of the castle with an initial velocity of 12m/s[42 degrees of above the horizontal]. The rock lands just on the far side of the castle's moat, at a level 9.5m below the initial level. Determine the rock's time of flight.
IsTim
 4 years ago
A medieval prince trapped in a castle wraps a message around a rock and throws it from the top of the castle with an initial velocity of 12m/s[42 degrees of above the horizontal]. The rock lands just on the far side of the castle's moat, at a level 9.5m below the initial level. Determine the rock's time of flight.

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hmmm...do u hav the answer..?

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.1I don't know how to get it, but the answer is 2.4s.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can u post it......it will be very helpful.....

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.1dw:1327933563123:dw IT's a lil rough

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0see for horizontal the time of flight is 2usintheta/g and fr vertical it is usintheta/g......so we need total T.O.F. so adding we get 3usintheta/g.....now substitute values..... 3*12*sin42/9.8 36*0.6691/9.8 24.0876/9.8 2.4s!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0time of flight[T.O.F]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0for horizontal motion and vertical motion[fr a projectile there always 2 types of motion horizontal and vertical]

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.1LEt me review your statement. It's a bit hard to understand....

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.1And how did you get those formulas>?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0u is initial velocity

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and those formulas can be derived

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but fr now just remember them

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.1Sorry. What do I derive those formulas from?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that is super long....just remember them or google it

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.1What should i be googling?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the derivation fr the formulas....range,max.height,t.o.f,etc/

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.1\[n _{1}\sin \theta_{1}=n _{2}\sin \theta_{2}\]

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.1I never learnt that formula in class. I'm doing an exam, so what is my best alternative?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wait lemme read it again.. i was just reading the comments :D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0nah.. i can't give a better solution sorry.. you need to know the dynamics of a parabolic tragectory!

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.1Bugger. Ok. I'll look thru my textbook more thoroughly then.

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.1There isn't an example with that formula in it. Argh

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.1The initial Velocity is the component of the original velocity right?

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.1wait, why is it multiplied by 3?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1327927068621:dw looking in the vertical direction,we have usin42 and acceleration g downwards at point A,final velocity for path till A becomes o so apply equations for uniform acceleration vu/t=a the SAME TIME is taken for stone to come to the same height so total time for 1st part is 2*usin42/g THIA GIVES US T1 the rock doesnt stop once it reaches the height from which the prince threw it does it? so we have a 2nd part in this problem dw:1327927440627:dw when it reaches point b(from where prince originally threw) at the same level, THE VELOCITY OF ROCK BECOMES SAME AS THE VELOCITY WITH WHICH PRINCE THREW that is usin42 now again apply equations for uniform acceleration dw:1327927625023:dw T1+T2 is the total time taken by rock! u have any doubt u may ask
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.