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Question Number 209161 by Spillover last updated on 02/Jul/24

Commented by klipto last updated on 03/Jul/24

$$$$$${\mathbf{F}}_{\mathbf{net}}=-({\mathbf{F}}_{\mathrm{g}}+\mathbf{P})$$$$\mathbf{ma}=-(\mathbf{mg}+0.2{\mathbf{v}}^2)$$$$\mathbf{a}=-(\mathbf{g}+\frac{{\mathbf{v}}^2}{5\mathbf{m}})$$$$\frac{\mathbf{dv}}{\mathbf{dt}}=-(\mathbf{g}+\frac{{\mathbf{v}}^2}{5\mathbf{m}})$$$$\mathbf{also}$$$$\mathbf{a}=\mathbf{v}\times \frac{\mathbf{dv}}{\mathbf{dh}}$$$$\therefore -(\mathbf{g}+\frac{{\mathbf{v}}^2}{5\mathbf{m}})=\frac{\mathbf{vdv}}{\mathbf{dh}}$$$$\frac{\mathbf{vdv}}{\mathbf{g}+\frac{{\mathbf{v}}^2}{5\mathbf{m}}}=-\mathbf{dh}$$$$\frac{\mathbf{v}}{\mathbf{g}+\frac{{\mathbf{v}}^2}{10}}\mathbf{dv}=-\mathbf{dh}$$$$\frac{\mathbf{v}}{\mathbf{g}+0.1{\mathbf{v}}^2}\mathbf{dv}=-\mathbf{dh}$$$${\int }_{\mathbf{u}}^{\mathbf{v}}\frac{\mathbf{v}}{\mathbf{g}+0.1{\mathbf{v}}^2}\mathbf{dv}=-\int \mathbf{dh},[\mathbf{u}=\mathbf{g}+0.1{\mathbf{v}}^2),\mathbf{dv}=\frac{\mathbf{du}}{0.2\mathbf{v}}$$$${\int }_{\mathbf{u}}^{\mathbf{v}}\frac{\mathbf{v}}{\mathbf{u}}\times \frac{\mathbf{du}}{0.2\mathbf{v}}=-\mathbf{h}$$$$\frac{1}{0.2}{\int }_{10}^0\frac{\mathbf{du}}{\mathbf{u}}=-\mathbf{h}$$$$5\mathbf{inu}{\mid }_{10}^0=-\mathbf{h}$$$$5[\mathbf{in}(\mathbf{g}+0.1{\mathbf{v}}^2){\mid }^0-\mathbf{in}(\mathbf{g}+0.1{\mathbf{v}}^2){\mid }^{10}=-\mathbf{h}$$$$5[\mathbf{in}\left(10\right)-\mathbf{in}(10+\frac{100}{10})]=5[\mathbf{in}10-\mathbf{in}20]=-\mathbf{h}$$$$-5\mathbf{ln}2=-\mathbf{h}$$$$\mathbf{h}=3.47$$$$\mathbf{klipto}-\mathbf{quanta}$$

Commented by Spillover last updated on 03/Jul/24

$$thankyou$$

Answered by mr W last updated on 02/Jul/24

$$ma=-mg-\frac{v^2}{5}$$$$a=-g-\frac{v^2}{5m}=\frac{dv}{dt}=v\frac{dv}{dh}$$$$-(g+\frac{v^2}{5m})=v\frac{dv}{dh}$$$$\frac{vdv}{g+\frac{v^2}{5m}}=-dh$$$$\int \frac{v^{2}dv}{g+\frac{v^2}{5m}}=-2\int dh$$$$\ln (g+\frac{v^2}{5m}){\mid }_u^v=-\frac{2h}{5m}$$$$\ln \frac{g+\frac{v^2}{5m}}{g+\frac{u^2}{5m}}=-\frac{2h}{5m}$$$$$$$$at{h}_{max}:v=0$$$$h_{max}=\frac{5m}{2}\ln (1+\frac{u^2}{5mg})$$$$=\frac{5\times 2}{2}\ln (1+\frac{{10}^2}{5\times 2\times 10})=3.47m$$

Commented by Spillover last updated on 02/Jul/24

$$yourright.thankyou$$

Answered by Spillover last updated on 02/Jul/24

Answered by Spillover last updated on 02/Jul/24

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