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Question Number 18268 by Tinkutara last updated on 17/Jul/17

$$\mathrm{A}\mathrm{balloon}\mathrm{starts}\mathrm{rising}\mathrm{from}\mathrm{the}\mathrm{surface}$$$$\mathrm{of}\mathrm{earth}.\mathrm{The}\mathrm{ascension}\mathrm{rate}\mathrm{is}\mathrm{constant}$$$$\mathrm{and}\mathrm{is}\mathrm{equal}\mathrm{to}{v}_0.\mathrm{Due}\mathrm{to}\mathrm{wind}\mathrm{the}$$$$\mathrm{balloon}\mathrm{gathers}\mathrm{horizontal}\mathrm{velocity}$$$$\mathrm{component}{v}_x=ay,\mathrm{where}a\mathrm{is}\mathrm{a}\mathrm{positive}$$$$\mathrm{constant}\mathrm{and}y\mathrm{is}\mathrm{the}\mathrm{height}\mathrm{of}\mathrm{ascent}.$$$$\mathrm{Find}$$$$\left(\mathrm{i}\right)\mathrm{The}\mathrm{horizontal}\mathrm{drift}\mathrm{of}\mathrm{the}\mathrm{balloon}$$$$x\left(y\right),$$$$\left(\mathrm{ii}\right)\mathrm{The}\mathrm{total},\mathrm{tangential}\mathrm{and}\mathrm{normal}$$$$\mathrm{accelerations}\mathrm{of}\mathrm{the}\mathrm{balloon}.$$

Answered by ajfour last updated on 16/Aug/17

Commented by Tinkutara last updated on 17/Aug/17

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

Commented by ajfour last updated on 16/Aug/17

$$y=v_{0}t,v_x=ay={av}_{0}t$$$$so{\int }_0^{x}dx={\int }_0^t{av}_{0}tdt=\frac{{av}_0{t}^2}{2}$$$$\Rightarrow x=\frac{{ay}^2}{2v_0}.$$$$a_y=0;{\overline{\mathit{a}}}_{\mathit{n}\mathit{e}\mathit{t}}=\left(\frac{{dv}_x}{dt}\right)\hat{i}={av}_0\hat{i}.$$$${\mathit{a}}_{\mathit{t}}=\mid {\overline{a}}_{nrt}\mid \cos \theta ={av}_0\frac{v_x}{\sqrt{v_x^2+v_y^2}}$$$${\mathit{a}}_{\mathit{t}}=\frac{{\mathit{a}\mathit{v}}_0\left({\mathit{a}\mathit{v}}_0\mathit{t}\right)}{\sqrt{{\left({\mathit{a}\mathit{v}}_0\mathit{t}\right)}^2+{\mathit{v}}_0^2}}=\frac{{\mathit{a}\mathit{v}}_0\mathit{t}}{\sqrt{{\mathit{a}}^2{\mathit{t}}^2+1}}.$$$${\mathit{a}}_{\mathit{N}}=\mid a_{net}\mid \sin \theta ={av}_0\frac{v_y}{\sqrt{v_x^2+v_y^2}}$$$${\mathit{a}}_{\mathit{N}}=\frac{{\mathit{a}\mathit{v}}_0}{\sqrt{{\mathit{a}}^2{\mathit{t}}^2+1}}.$$

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