Question Number 19890 by NECC last updated on 17/Aug/17
$${A}\:{man}\:{on}\:{top}\:{of}\:{a}\:{tower}\:{of}\:{height} \\ $$$$\mathrm{35}{m}\:{throws}\:{a}\:{stone}\:{vertically} \\ $$$${upwards}\:{with}\:{a}\:{speed}\:{of}\:\mathrm{14}{m}/{s}. \\ $$$${Find}: \\ $$$$\left({i}\right){the}\:{height}\:{above}\:{the}\:{ground}, \\ $$$${reached}\:{by}\:{the}\:{stone}. \\ $$$$\left({ii}\right){the}\:{speed}\:{of}\:{the}\:{stone},{when} \\ $$$${it}\:{reaches}\:{the}\:{ground}. \\ $$
Answered by mrW1 last updated on 17/Aug/17
$$\left(\mathrm{i}\right) \\ $$$$\mathrm{mg}\Delta\mathrm{h}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{mv}_{\mathrm{0}} ^{\mathrm{2}} \\ $$$$\Delta\mathrm{h}=\frac{\mathrm{v}_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{2g}}=\frac{\mathrm{14}^{\mathrm{2}} }{\mathrm{2}×\mathrm{9}.\mathrm{81}}\approx\mathrm{10}\:\mathrm{m} \\ $$$$\mathrm{h}=\mathrm{h}_{\mathrm{T}} +\Delta\mathrm{h}=\mathrm{35}+\mathrm{10}=\mathrm{45}\:\mathrm{m} \\ $$$$\left(\mathrm{ii}\right) \\ $$$$\mathrm{mgh}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{mv}^{\mathrm{2}} \\ $$$$\mathrm{v}=\sqrt{\mathrm{2gh}}=\sqrt{\mathrm{2}×\mathrm{9}.\mathrm{81}×\mathrm{45}}\approx\mathrm{29}.\mathrm{7}\:\mathrm{m}/\mathrm{s} \\ $$