projectile question??? (1 Viewer)

freaking_out

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how did u guys do the projectile question...i. e the last bit...i couldn't do it, i got this huge equation which i had to solve to get d...so yeah, i ran out of time. :eek:
 

nerdd

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nope couldn't do it?
I had v^2 on g in my expression tho after i differentiated and stuff so i just said yeah...... that's one of the solutions.. AS REQUIRED haha
 

Neon-Frog

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I used pythagora's.

The max height was the equation u derived in part i), and the x value was d/2. then u pythagoras and sub other crap in.. I didnt get it all coz i ran out of time .. but i got the raw equations down.
 

Constip8edSkunk

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ran out of time didnt finish... dont think many ppl did....
heres what i got up 2
show that pi/4 maximise distance
use eqn in i) to get an expression for H-S
use eqn in ii) to get an expression for d
fiddle around a bit, u get the 2nd eqn... thats if the parabola wont touch the ceiling...

this will change if the velocity is faster though... and i didnt finish this bit...im guessing you have to make an expression for alpha and sub it in.
 

iambored

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iwas not happy we didn't have to prove the 2 top equations! i would have got more marks!
 

Constip8edSkunk

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got the other eqn, the 1st 1, original guess was wrong:D

d=2v^2sin@cos@/g
d^2=4v^2sin^2@(1-sin^2@)/g^2
=16 [ v^2sin^2@/2g] [v^2/2g - v^2sin^2@/2g]
=16 [H-S][v^2/2g - [H-S]]
d=4sqrt{ [H-S][v^2/2g] - [H-S]^2}

EDIT: oh yeah this is when @=/= pi/4 as if it was, it would hit the ceiling, u get the inequality condition from the 1st part which some how lead to the 2nd eqn lol

if only i had time in the exam... or had not spent 30 min bashing the bearing question lol
 
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Jumbo Cactuar

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yeah I managed to prove the smaller part for @ >= pi/4 but not enough inspiration or time for the bastard for @ <= pi/4

That exam was a walk in the park, if you are into that sorta thing... (what?)
 

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