ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: 02315 on October 30, 2010, 09:04:06 pm
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Just to confirm, was this taken out of the course last year?
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Just to confirm, was this taken out of the course last year?
If it did, then would it not have said on the VCAA site?
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The heck is projectile motion?
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question 19 on the MC 2009 exam 2, wouldnt you consider that projectile motion, maybe just because im physics/spesh
edit: vcaa 2009 exam 2
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question 19 on the MC 2009 exam 2, wouldnt you consider that projectile motion, maybe just because im physics/spesh
oh... it is? isn't that just constant acceleration formula shiz?
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i dno how would i do it then
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Well using the formulas, you have: taking down as positive and up as negative
a = g, as the cricket ball's mass is negligible
u = -20sin(45), as you are only taking vertical into consideration here
v = 20sin(45), since it came from ground, when it comes back to the ground it is at the same velocity with positive sign since downwards
t = ?
v = u + at
t = (v-u)/a
= 40sin(45)/g
= 40 root(2) / g*2
= 20root(2)/g
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ohh i see, i don't really understand why v = 20sin(45) though, as in the value part of it
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Well if you just picture the path of the cricket ball, it moves in a parabolic shape, symmetrical about the point where it changes direction of motion, so wouldn't where it leaves the ground be the same velocity as when it hits the ground again?
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I believe it is the same velocity 'up' and 'down'
we only need to take the vertical velocity (horizontal doesn't matter)
ie. the time it takes from ground to highest point = the time it takes from highest point to ground
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guess it makes sense if you put it like that, thanks for the help :)