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VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Physics => Topic started by: cranberry on June 11, 2011, 05:47:16 pm

Title: VCAA 2008 question 7
Post by: cranberry on June 11, 2011, 05:47:16 pm: solutions aren't clear, am i missing something?...see attached

i got total time of 3 seconds, V(horizontal) as 24m per s, t (from max height to board) of 1.2s; used x = ut + .5at^2 and got 7.2m.

answer's 9m - see attached.

any help?

ty
Title: Re: VCAA 2008 question 7
Post by: onur369 on June 11, 2011, 05:55:44 pm: Okay... Now your mind might explode whilst I explain this.

We have RangeMax we have initial speed. Thus we do RangeMax = ux x tflight
make tflight the subject. Find ux, ux is found by uxcostheta which in this case is 24.
72/24 = 3 <=== TIME

use this formula: h=uy x t x 5t^2

we need uy thus we do uy=usintheta in this case 30xsin36.9 which is 18
h=uyt x 5t^2
= 18x3 - 5(3^2)
=54-45
=9m
Title: Re: VCAA 2008 question 7
Post by: Lasercookie on June 11, 2011, 06:00:22 pm: Split into horizontal/vertical components:
(Horizontal)
$u=30cos36.9°, x=72$
$t=x/u$
$t=72/30cos36.9°= 3s$ (don't round off in your calculator)

(Vertical)
$u=30sin36.9°, a=-10, t=3$
$x=ut+1/2at^2$
$x=(30sin36.9)(3)+1/2(-10)(3)^2$
$x=9m$

Edit: Beaten, but I prefer to not use derived formulas and just do things the regular method by splitting into horizontal/vertical components. That way I can be sure to be approaching the situation described correctly. The time that it takes to complete the question is about the same anyway.
Title: Re: VCAA 2008 question 7
Post by: cranberry on June 11, 2011, 07:00:59 pm: yeh that gets 9m, but why have you put in a value for u when its from max height (ie, u should be zero no?)
and a time of 3s from max height?? or have you done something different?
Title: Re: VCAA 2008 question 7
Post by: cranberry on June 11, 2011, 07:27:43 pm: bump^^.....and just to let you know what i did, i found the time from start to max height (1.8 seconds) and subtracted it from the total time (3 seconds) = 1.2 seconds - the time from max height to the board?!?! gotta be right no?
Title: Re: VCAA 2008 question 7
Post by: ttn on June 11, 2011, 07:46:09 pm: If you're going to do it like that, you will have to minus that value away from the max height since it fell 7.2 m, not rise.

v^2 = u^2 + 2ad
d = 324/20
d = 16.2
max h = 16.2

h billboard = max h - lost h
= 16.2 - 7.2
= 9 m
Title: Re: VCAA 2008 question 7
Post by: cranberry on June 11, 2011, 07:52:59 pm: ahhhhhhhhhhhh yes. thankyou ttn!