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Question-224406




Question Number 224406 by Tawa11 last updated on 08/Sep/25
Answered by mr W last updated on 09/Sep/25
mur sin θ_0 =mUr  ⇒u sin θ_0 =U  ((mu^2 )/2)=((mU^2 )/2)+mgr cos θ_0   (u^2 /2)=((u^2 sin^2  θ_0 )/2)+gr cos θ_0   u^2  cos θ_0 =2gr  ⇒u=(√((2gr)/(cos θ_0 ))) ✓
$${mur}\:\mathrm{sin}\:\theta_{\mathrm{0}} ={mUr} \\ $$$$\Rightarrow{u}\:\mathrm{sin}\:\theta_{\mathrm{0}} ={U} \\ $$$$\frac{{mu}^{\mathrm{2}} }{\mathrm{2}}=\frac{{mU}^{\mathrm{2}} }{\mathrm{2}}+{mgr}\:\mathrm{cos}\:\theta_{\mathrm{0}} \\ $$$$\frac{{u}^{\mathrm{2}} }{\mathrm{2}}=\frac{{u}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta_{\mathrm{0}} }{\mathrm{2}}+{gr}\:\mathrm{cos}\:\theta_{\mathrm{0}} \\ $$$${u}^{\mathrm{2}} \:\mathrm{cos}\:\theta_{\mathrm{0}} =\mathrm{2}{gr} \\ $$$$\Rightarrow{u}=\sqrt{\frac{\mathrm{2}{gr}}{\mathrm{cos}\:\theta_{\mathrm{0}} }}\:\checkmark \\ $$
Commented by mr W last updated on 10/Sep/25
Commented by fantastic last updated on 09/Sep/25
why were you inactive for  some days??
$${why}\:{were}\:{you}\:{inactive}\:{for} \\ $$$${some}\:{days}?? \\ $$
Commented by Tawa11 last updated on 13/Sep/25
Thanks sir, I appreciate.
$$\mathrm{Thanks}\:\mathrm{sir},\:\mathrm{I}\:\mathrm{appreciate}. \\ $$
Commented by mr W last updated on 13/Sep/25
did you comfirm that the answer is  correct?
$${did}\:{you}\:{comfirm}\:{that}\:{the}\:{answer}\:{is} \\ $$$${correct}? \\ $$
Commented by Tawa11 last updated on 14/Sep/25
Yes sir.  they wrote     (√(2gr cosθ_0 ))
$$\mathrm{Yes}\:\mathrm{sir}. \\ $$$$\mathrm{they}\:\mathrm{wrote}\:\:\:\:\:\sqrt{\mathrm{2gr}\:\mathrm{cos}\theta_{\mathrm{0}} } \\ $$
Commented by mr W last updated on 14/Sep/25
what do you think is right,  (√((2gr)/(cos θ_0 ))) or (√(2gr cos θ_0 )) ?
$${what}\:{do}\:{you}\:{think}\:{is}\:{right}, \\ $$$$\sqrt{\frac{\mathrm{2}{gr}}{\mathrm{cos}\:\theta_{\mathrm{0}} }}\:{or}\:\sqrt{\mathrm{2}{gr}\:\mathrm{cos}\:\theta_{\mathrm{0}} }\:? \\ $$
Commented by fantastic last updated on 07/Oct/25
I also got  u=(√((2gr)/(cos θ_0 )))
$${I}\:{also}\:{got} \\ $$$${u}=\sqrt{\frac{\mathrm{2}{gr}}{\mathrm{cos}\:\theta_{\mathrm{0}} }} \\ $$

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