## 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.

1. anonymous

2. IsTim

I don't know how to get it, but the answer is 2.4s.

3. anonymous

ok...wait.....

4. anonymous

is a diagram given?

5. IsTim

Yes!

6. anonymous

can u post it......it will be very helpful.....

7. IsTim

|dw:1327933563123:dw| IT's a lil rough

8. anonymous

i am gettin 1.6

9. IsTim

?

10. anonymous

i got 2.4!!!!!!

11. anonymous

istim u there?

12. IsTim

How?

13. anonymous

see 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!

14. IsTim

fr? TOF?

15. anonymous

time of flight[T.O.F]

16. IsTim

and fr?

17. anonymous

for horizontal motion and vertical motion[fr a projectile there always 2 types of motion horizontal and vertical]

18. anonymous

ok...got it?

19. IsTim

LEt me review your statement. It's a bit hard to understand....

20. IsTim

wats u in 2usintheta/g

21. IsTim

And how did you get those formulas>?

22. anonymous

u is initial velocity

23. anonymous

and those formulas can be derived

24. anonymous

but fr now just remember them

25. IsTim

Sorry. What do I derive those formulas from?

26. anonymous

that is super long....just remember them or google it

27. IsTim

What should i be googling?

28. anonymous

the derivation fr the formulas....range,max.height,t.o.f,etc/

29. IsTim

$n _{1}\sin \theta_{1}=n _{2}\sin \theta_{2}$

30. IsTim

is it that one?

31. IsTim

hello?

32. anonymous
33. anonymous

ok?

34. IsTim

I never learnt that formula in class. I'm doing an exam, so what is my best alternative?

35. anonymous

idk....

36. IsTim

Any opinions mashy?

37. anonymous

38. IsTim

k

39. anonymous

nah.. i can't give a better solution sorry.. you need to know the dynamics of a parabolic tragectory!

40. IsTim

Bugger. Ok. I'll look thru my textbook more thoroughly then.

41. IsTim

There isn't an example with that formula in it. Argh

42. IsTim

The initial Velocity is the component of the original velocity right?

43. IsTim

wait, why is it multiplied by 3?

44. anonymous

45. IsTim

eh. which one?

46. anonymous

|dw: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 v-u/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