How to do Q27 proj. motion. 2012 HSC paper physics? (1 Viewer)

nightweaver066 · Jan 14, 2013

Good ol' angry birds :')

$u_y = vsin60=\frac{\sqrt{3}}{2}v$

$u_x = vcos60 = \frac{1}{2}v$

$\Delta x = u_x t$

$45 = \frac{1}{2}vt$

$\Delta y = u_y t + \frac{1}{2}a_yt^2$

$34 = \frac{\sqrt{3}}{2}vt - 4.9t^2$

$34 = 45\sqrt{3}-4.9t^2$

$t = \sqrt{\frac{45\sqrt{3}-34}{4.9}$

$\approx 2.99$

$v = \frac{90}{t}$

$= \frac{90}{2.99}$

$= 30m/s \; $(1 s.f.)$

darkfenrir · Jan 15, 2013

Thankyou friend, I didn't pay attention in class when projectile motion was being done lol.

D94 · Jan 15, 2013

One thing you should note is that given height is not necessarily the maximum height of the projectile. Don't be fooled by the diagram, just use the data given - some students assumed it was the maximum height and whilst their final answer was close to the real answer, their method/solution was completely incorrect.

darkfenrir · Jan 15, 2013

D94 said:
One thing you should note is that given height is not necessarily the maximum height of the projectile. Don't be fooled by the diagram, just use the data given - some students assumed it was the maximum height and whilst their final answer was close to the real answer, their method/solution was completely incorrect.

But it says "strike the base of the bulding" so isnt the total change in y the value listed? Oh wait i see what you mean. So how do you work out maximum height?

How to do Q27 proj. motion. 2012 HSC paper physics? (1 Viewer)

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