|
|
|
STEP II 1997 question 11 solutionTSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP II 1997 question 11 solution At the ground,
QED
At the net,
Substitute in v^2 from the first part, giving:
A little bit of rearranging to get tan(alpha) as the subject gives:
QED
v^2 is always positive so
tan(alpha) is always positive too (the question says that the ball is hit downward, so 0<alpha<90). So:
Solution by ad absurdum |
|
|||||||||||||||
![\displaystyle v^2 = \frac{g(a+b)^2 \sec^2\alpha}{2[H-(a+b)\tan\alpha]}](http://www.thestudentroom.co.uk/latexrender/pictures/23/2317cee94add30fa9b957cfbf33af0bf.png "\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]}](http://www.thestudentroom.co.uk/latexrender/pictures/d2/d26e721cc7f4f309b08397927b1f7129.png "\displaystyle v^2 = \frac{g(a+b)^2(1+\tan^2\alpha}{2[h-(a+b)\tan\alpha]}")
![\displaystyle H-a\tan\alpha - \frac{a^2[h-(a+b)\tan\alpha]}{(a+b)^2}=h](http://www.thestudentroom.co.uk/latexrender/pictures/16/16f2d1a4d3392bfdb3179d586669a073.png "\displaystyle H-a\tan\alpha - \frac{a^2[h-(a+b)\tan\alpha]}{(a+b)^2}=h")
