STEP II 1992 question 11 solution

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STEP II 1992 question 11 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP II 1992 question 11 solution

if the purse scrapes the edge of the cliff, it reaches its max point at 0.5d from edge of cliff. vertical movement: u=VsinA, v=0, a=-g $\implies s = \dfrac{(V\sin A)^2}{2g}$

horizontal movement after max point: v=VcosA, s=0.5d $\implies t = \dfrac{d}{2V\cos A}$ vertical movement after max point: $s=\dfrac{(V\sin A)^2}{2g} - h$ , u=0, a=g, $t = \dfrac{d}{2V\cos A}$

$\displaystyle \frac{(V\sin A)^2}{2g} - h = \frac{0.5gd}{2V\cos A}$

$\displaystyle \implies 8V^4\sin^4A - 8V^2(2gh+V^2)\sin^2A + 2g(gd^2+8V^2h) = 0\\ \\ \sin^2A = \frac{8V^2(2gh+V^2) \pm \sqrt{}(64V^4(V^4+4gh+4g^2h^2) - 64gV^4(gd^2+8V^2h))}{16V^4}$