STEP II 1997 question 11 solution

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STEP II 1997 question 11 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP II 1997 question 11 solution

At the ground, $\displaystyle vt\cos\alpha = a+b$ , $\displaystyle H-vt\sin\alpha - \frac{1}{2}gt^2=0$ . Substitute in t from horizontal equation into vertical one giving:

$\displaystyle H-(a+b)\tan\alpha - \frac{g(1+b)^2}{2v^2\cos^2\alpha} = 0$

$\displaystyle v^2 = \frac{g(a+b)^2 \sec^2\alpha}{2[H-(a+b)\tan\alpha]}$

$\displaystyle v^2 = \frac{g(a+b)^2(1+\tan^2\alpha}{2[h-(a+b)\tan\alpha]}$

QED

At the net, $\displaystyle vt\cos\alpha=a$ and $\displaystyle H-vt\sin\alpha - \frac{1}{2}gt^2=0$ . Again, subsititute in t from the horizontal equation, giving:

$\displaystyle H-a\tan\alpha - \frac{ga^2\sec^2\alpha}{2v^2}=h$

Substitute in v^2 from the first part, giving:

$\displaystyle H-a\tan\alpha - \frac{a^2[h-(a+b)\tan\alpha]}{(a+b)^2}=h$

A little bit of rearranging to get tan(alpha) as the subject gives:

$\displaystyle \tan\alpha = \frac{2a+b}{a(a+b)}H -\frac{a+b}{ab}h$

QED

v^2 is always positive so $\displaystyle H>(a+b)\tan\alpha$ (from the denominator of the first part). Substitute tan(alpha) from the previous section to get: