Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

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

A balloon starts rising from the surface  of earth. The ascension rate is constant  and is equal to v_0 . Due to wind the  balloon gathers horizontal velocity  component v_x  = ay, where a is a positive  constant and y is the height of ascent.  Find  (i) The horizontal drift of the balloon  x(y),  (ii) The total, tangential and normal  accelerations of the balloon.

$$\mathrm{A}\:\mathrm{balloon}\:\mathrm{starts}\:\mathrm{rising}\:\mathrm{from}\:\mathrm{the}\:\mathrm{surface} \\ $$$$\mathrm{of}\:\mathrm{earth}.\:\mathrm{The}\:\mathrm{ascension}\:\mathrm{rate}\:\mathrm{is}\:\mathrm{constant} \\ $$$$\mathrm{and}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:{v}_{\mathrm{0}} .\:\mathrm{Due}\:\mathrm{to}\:\mathrm{wind}\:\mathrm{the} \\ $$$$\mathrm{balloon}\:\mathrm{gathers}\:\mathrm{horizontal}\:\mathrm{velocity} \\ $$$$\mathrm{component}\:{v}_{{x}} \:=\:{ay},\:\mathrm{where}\:{a}\:\mathrm{is}\:\mathrm{a}\:\mathrm{positive} \\ $$$$\mathrm{constant}\:\mathrm{and}\:{y}\:\mathrm{is}\:\mathrm{the}\:\mathrm{height}\:\mathrm{of}\:\mathrm{ascent}. \\ $$$$\mathrm{Find} \\ $$$$\left(\mathrm{i}\right)\:\mathrm{The}\:\mathrm{horizontal}\:\mathrm{drift}\:\mathrm{of}\:\mathrm{the}\:\mathrm{balloon} \\ $$$${x}\left({y}\right), \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{The}\:\mathrm{total},\:\mathrm{tangential}\:\mathrm{and}\:\mathrm{normal} \\ $$$$\mathrm{accelerations}\:\mathrm{of}\:\mathrm{the}\:\mathrm{balloon}. \\ $$

Answered by ajfour last updated on 16/Aug/17

Commented by Tinkutara last updated on 17/Aug/17

Thank you very much Sir!

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

Commented by ajfour last updated on 16/Aug/17

  y=v_0 t  ,  v_x =ay=av_0 t   so    ∫_0 ^(  x) dx =∫_0 ^(  t) av_0 tdt =((av_0 t^2 )/2)      ⇒     x=((ay^2 )/(2v_0 ))   .         a_y =0 ;    a_(net) ^� =((dv_x /dt))i^�  =av_0 i^�  .        a_t =∣a_(nrt) ^� ∣cos θ  =av_0 (v_x /(√(v_x ^2 +v_y ^2 )))     a_t  =((av_0 (av_0 t))/(√((av_0 t)^2 +v_0 ^2 ))) =((av_0 t)/(√(a^2 t^2 +1))) .    a_N =∣a_(net) ∣sin θ =av_0 (v_y /(√(v_x ^2 +v_y ^2 )))     a_N =((av_0 )/(√(a^2 t^2 +1)))  .

$$\:\:{y}={v}_{\mathrm{0}} {t}\:\:,\:\:{v}_{{x}} ={ay}={av}_{\mathrm{0}} {t} \\ $$$$\:{so}\:\:\:\:\int_{\mathrm{0}} ^{\:\:{x}} {dx}\:=\int_{\mathrm{0}} ^{\:\:{t}} {av}_{\mathrm{0}} {tdt}\:=\frac{{av}_{\mathrm{0}} {t}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\:\:\:\:\Rightarrow\:\:\:\:\:{x}=\frac{{ay}^{\mathrm{2}} }{\mathrm{2}{v}_{\mathrm{0}} }\:\:\:. \\ $$$$\:\:\:\:\:\:\:{a}_{{y}} =\mathrm{0}\:;\:\:\:\:\bar {\boldsymbol{{a}}}_{\boldsymbol{{net}}} =\left(\frac{{dv}_{{x}} }{{dt}}\right)\hat {{i}}\:={av}_{\mathrm{0}} \hat {{i}}\:. \\ $$$$\:\:\:\:\:\:\boldsymbol{{a}}_{\boldsymbol{{t}}} =\mid\bar {{a}}_{{nrt}} \mid\mathrm{cos}\:\theta\:\:={av}_{\mathrm{0}} \frac{{v}_{{x}} }{\sqrt{{v}_{{x}} ^{\mathrm{2}} +{v}_{{y}} ^{\mathrm{2}} }}\: \\ $$$$\:\:\boldsymbol{{a}}_{\boldsymbol{{t}}} \:=\frac{\boldsymbol{{av}}_{\mathrm{0}} \left(\boldsymbol{{av}}_{\mathrm{0}} \boldsymbol{{t}}\right)}{\sqrt{\left(\boldsymbol{{av}}_{\mathrm{0}} \boldsymbol{{t}}\right)^{\mathrm{2}} +\boldsymbol{{v}}_{\mathrm{0}} ^{\mathrm{2}} }}\:=\frac{\boldsymbol{{av}}_{\mathrm{0}} \boldsymbol{{t}}}{\sqrt{\boldsymbol{{a}}^{\mathrm{2}} \boldsymbol{{t}}^{\mathrm{2}} +\mathrm{1}}}\:. \\ $$$$\:\:\boldsymbol{{a}}_{\boldsymbol{{N}}} =\mid{a}_{{net}} \mid\mathrm{sin}\:\theta\:={av}_{\mathrm{0}} \frac{{v}_{{y}} }{\sqrt{{v}_{{x}} ^{\mathrm{2}} +{v}_{{y}} ^{\mathrm{2}} }}\: \\ $$$$\:\:\boldsymbol{{a}}_{\boldsymbol{{N}}} =\frac{\boldsymbol{{av}}_{\mathrm{0}} }{\sqrt{\boldsymbol{{a}}^{\mathrm{2}} \boldsymbol{{t}}^{\mathrm{2}} +\mathrm{1}}}\:\:. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com