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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:

\displaystyle H<\frac{h(a+b)^2}{b(a-b)}


tan(alpha) is always positive too (the question says that the ball is hit downward, so 0<alpha<90). So:

\displaystyle H\frac{2a+b}{a(a+b)}>\frac{a+b}{ab}h

\displaystyle H>\frac{(a+b)^2}{b(2a+b)}h

Solution by ad absurdum

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