A-wooden-block-of-mass-10-gm-is-dropped-from-the-top-of-a-tower-100-m-high-Simultaneously-a-bullet-of-mass-10-gm-is-fired-from-the-foot-of-the-tower-vertically-upwards-with-a-velocity-of-100-m-sec-

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Question Number 23520 by Tinkutara last updated on 01/Nov/17

$$\mathrm{A}\mathrm{wooden}\mathrm{block}\mathrm{of}\mathrm{mass}10\mathrm{gm}\mathrm{is}$$$$\mathrm{dropped}\mathrm{from}\mathrm{the}\mathrm{top}\mathrm{of}\mathrm{a}\mathrm{tower}100\mathrm{m}$$$$\mathrm{high}.\mathrm{Simultaneously},\mathrm{a}\mathrm{bullet}\mathrm{of}\mathrm{mass}$$$$10\mathrm{gm}\mathrm{is}\mathrm{fired}\mathrm{from}\mathrm{the}\mathrm{foot}\mathrm{of}\mathrm{the}\mathrm{tower}$$$$\mathrm{vertically}\mathrm{upwards}\mathrm{with}\mathrm{a}\mathrm{velocity}\mathrm{of}$$$$100\mathrm{m}/\sec .\mathrm{If}\mathrm{the}\mathrm{bullet}\mathrm{is}\mathrm{embedded}\mathrm{in}$$$$\mathrm{it},\mathrm{how}\mathrm{high}\mathrm{will}\mathrm{it}\mathrm{rise}\mathrm{above}\mathrm{the}\mathrm{tower}$$$$\mathrm{before}\mathrm{it}\mathrm{starts}\mathrm{falling}?(\mathrm{Consider}g=$$$$10\mathrm{m}/{\sec }^2)$$

Commented by Tinkutara last updated on 01/Nov/17

Commented by mrW1 last updated on 01/Nov/17

$${\mathrm{h}}_{\mathrm{w}}=\mathrm{h}-\frac{1}{2}{\mathrm{gt}}^2$$$${\mathrm{h}}_{\mathrm{b}}=\mathrm{ut}-\frac{1}{2}{\mathrm{gt}}^2$$$${\mathrm{h}}_{\mathrm{w}}={\mathrm{h}}_{\mathrm{b}}$$$$\mathrm{h}-\frac{1}{2}{\mathrm{gt}}^2=\mathrm{ut}-\frac{1}{2}{\mathrm{gt}}^2$$$$\Rightarrow \mathrm{t}=\frac{\mathrm{h}}{\mathrm{u}}=\frac{100}{100}=1\sec$$$$\Rightarrow {\mathrm{h}}_1=100-\frac{1}{2}\times 10\times 1^2=95\mathrm{m}$$$${\mathrm{v}}_{\mathrm{wi}}=\mathrm{gt}=10\mathrm{m}/\mathrm{s}(\downarrow )$$$${\mathrm{v}}_{\mathrm{bi}}=\mathrm{u}-\mathrm{gt}=100-10=90\mathrm{m}/\mathrm{s}(\uparrow )$$$${\mathrm{m}}_{\mathrm{b}}{\mathrm{v}}_{\mathrm{bi}}-{\mathrm{m}}_{\mathrm{w}}{\mathrm{v}}_{\mathrm{wi}}=({\mathrm{m}}_{\mathrm{b}}+{\mathrm{m}}_{\mathrm{w}}){\mathrm{v}}_{\mathrm{f}}$$$$\Rightarrow {\mathrm{v}}_{\mathrm{f}}=\frac{10\times 90-10\times 10}{20}=40\mathrm{m}/\mathrm{s}(\uparrow )$$$$\mathrm{m}={\mathrm{m}}_{\mathrm{b}}+{\mathrm{m}}_{\mathrm{w}}$$$${\mathrm{mgh}}_1+\frac{1}{2}{\mathrm{mv}}_{\mathrm{f}}^2={\mathrm{mgh}}_2$$$$\Rightarrow {\mathrm{h}}_2={\mathrm{h}}_1+\frac{{\mathrm{v}}_{\mathrm{f}}^2}{2\mathrm{g}}=95+\frac{{40}^2}{2\times 10}=175\mathrm{m}$$$$\Rightarrow \mathrm{\Delta }\mathrm{h}={\mathrm{h}}_2-\mathrm{h}=175-100=75\mathrm{m}$$$$\mathrm{i}.\mathrm{e}.\mathrm{they}\mathrm{start}\mathrm{falling}75\mathrm{m}\mathrm{above}\mathrm{the}\mathrm{tower}.$$

Commented by Tinkutara last updated on 01/Nov/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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