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Average Boreduser

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So q14 c) for 2022 hsc math ex 1, WHY DO WE MAXIMISE THE d if we've established that the max range is when theta= pi/4? shouldn't it mean we should sub in 45* into the trig functions and state that d<4u^2/g-4u^2/sqrt2*g? It's not the right answer but I can't see why we negate the fact that the max range would be when theta= 45* and instead maximise the d///
 

Hughmaster

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I understand where you're coming from. Due to the time taken to attain this maximum range (with theta = 45) you have to realise that the target is ALSO moving and will have its own relative position given the time of flight for both particles. So the heuristic of plugging in theta = 45 no longer works.
 

coolcat6778

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So q14 c) for 2022 hsc math ex 1, WHY DO WE MAXIMISE THE d if we've established that the max range is when theta= pi/4? shouldn't it mean we should sub in 45* into the trig functions and state that d<4u^2/g-4u^2/sqrt2*g? It's not the right answer but I can't see why we negate the fact that the max range would be when theta= 45* and instead maximise the d///
Type in latex.
 

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