Projectile Motion?

[02315]

Just to confirm, was this taken out of the course last year?

[Souljette_93]

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?

[98.40_for_sure]

The heck is projectile motion?

[02315]

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

[98.40_for_sure]

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?

[02315]

i dno how would i do it then

[98.40_for_sure]

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

[02315]

ohh i see, i don't really understand why v = 20sin(45) though, as in the value part of it

[98.40_for_sure]

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?

[jasoN-]

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

[02315]

guess it makes sense if you put it like that, thanks for the help