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HSC Stuff => HSC Maths Stuff => HSC Subjects + Help => HSC Mathematics Extension 1 => Topic started by: helkit98 on September 29, 2016, 01:30:26 pm

Title: Mathematics Extension 1 - HSC 2001 Q4 b) projectiles
Post by: helkit98 on September 29, 2016, 01:30:26 pm: The answer is 200m/s, but I am not sure how to work it out...
Title: Re: Mathematics Extension 1 - HSC 2001 Q4 b) projectiles
Post by: RuiAce on September 29, 2016, 01:49:37 pm: Quote from: helkit98 on September 29, 2016, 01:30:26 pm
The answer is 200m/s, but I am not sure how to work it out...
$\text{Because we don't know the height, we need to use the angle of impact.}\\ \text{First note that when }t=T, \, x=4000\text{ where }T\text{ is the time of flight.}\\ \therefore 4000=VT \implies \boxed{T=\frac{4000}{V}}$
$\text{To consider the angle of impact, we need }\frac{dy}{dx}.\text{ Now:}\\ \begin{align*}\frac{dx}{dt}&=V\\ \frac{dy}{dt}&=-10t\\ \therefore \frac{dy}{dx}&=\frac{dy}{dt}\,\frac{dt}{dx}\tag{!}\\ &=-\frac{10t}{V} \end{align*}$
$\text{So upon landing, i.e. when }t=T\text{ we have}\\ \boxed{\frac{dy}{dx}=-\frac{40000}{V^2}}$
$\text{But the angle of landing tells us something.}\\ \text{The angle of landing allows us to use the result }m=\tan \theta\\ \text{Here, }\theta=-45^\circ$
$\therefore m=\tan (-45^\circ) = -1 = \frac{dy}{dx}\\ \text{because the derivative is the slope of the tangent.}$
$\text{So upon equating:}\\ -1=-\frac{40000}{V^2}\implies \boxed{V=200}$
$\boxed{\text{You should definitely understand why }m=\tan\theta = \frac{dy}{dx}}\\ \text{The chain rule is quite common for angle of landing.}\\ \text{Another approach is parametric differentiation: } \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$