jxballistic
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Hi guys I've been stuck on question 12 of ex25d for a while now. Can anyone show me how it's done?
Projectile motion question from Fitzpatrick (1 Viewer)
- Thread starter jxballistic
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After the usual standard derivations:
 = Vtcos \alpha )
 = Vtsin \alpha - 5t^2 )
 = Vsin \alpha \times \frac {7}{Vcos \alpha} - 5(\frac {7}{Vcos \alpha})^2 = 7 \implies 7(tan \alpha -1) = \frac {5 \cdot 7\cdot 7}{V^2 \cdot cos^2 \alpha})
 )
Similarly, using T2:
}{4} )
Equating the 2 will give you:
I've done the 2nd part - but I'm too slow at typing out in LaTeX. The basic approach is the same. Instead of height 7, replace with 'h'. Instead of 1st horizontal distance 7, use 'b'and replace 14 with c. Fiddle around with the 2 distinct but equivalent expressions for 5/( V^2 x cos^2 @) you'll get the desired results.
Similarly, using T2:
Equating the 2 will give you:
I've done the 2nd part - but I'm too slow at typing out in LaTeX. The basic approach is the same. Instead of height 7, replace with 'h'. Instead of 1st horizontal distance 7, use 'b'and replace 14 with c. Fiddle around with the 2 distinct but equivalent expressions for 5/( V^2 x cos^2 @) you'll get the desired results.
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Carrotsticks
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Sydney Grammar 2011 Q7 (b)
jxballistic
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Thanks drongoski, that helped a lot 
. Carrotsticks was this question in sydney grammars paper?
. Carrotsticks was this question in sydney grammars paper?
Carrotsticks
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Yep it was. There are a couple of ways to prove that property, I've seen it quite a few times now.Thanks drongoski, that helped a lot