Question

Shots are fired simultaneously from the top and the bottom of a vertical cliff with elevation ${37}^o$ and ${53}^o$ respectively, strike an object at the ground simultaneously. If the horizontal distance of the object from the cliff is $60$m, find the height of the cliff.

Answer 1

Coordinates of P wrt O and A are (s, n) and (s, -m).

$\alpha ={37}^o,\beta ={53}^o$

The height of the cliff $\Rightarrow h=(m+n)$

$y=x\tan \alpha -\frac{g{x}^2}{2x^2{\cos }^2\alpha }$

$\therefore -m=s\tan \alpha -\frac{g{s}^2}{2u^2{\cos }^2\alpha }$

$\therefore n=s\tan \beta -\frac{g{s}^2}{2v^2{\cos }^2\beta }$

and $(m+n)=s(\tan \beta -\tan \alpha )$

Because $u\cos \alpha =v\cos \beta$, as both the shots reach at point P simultaneously. This is the only possibility, if horizontal components of velocities are equal.

$\therefore n=s(\tan \beta -\tan \alpha )-s(\frac{1}{\cot \beta }-\frac{1}{\cot \alpha })=\frac{s(\cot \alpha -\cot \beta )}{cot\alpha \cos \beta }$

$\Rightarrow 60(\cot {37}^o-\cot {53}^o)/\cot {37}^o.\cot {53}^o=35m$