## Physics (10th Edition)

Let down be the negative direction $v^{2}=u^{2}+2ax$ $u=±\sqrt{ ((-27)^{2}-2(-9.8)(-15))}$ $u=-\sqrt{ (435)}$ since the gun was fired downwards To find maximum height above the cliff edge: $v^{2}=u^{2}+2ax$ $x=\frac{(0)^{2}-435}{2(-9.8)}\approx22$ m
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