# 102 - Kinematics Questions Answers

At t = 0, a stone is dropped into a well in which the level of water is h below the top of the well. If v is velocity of sound, the time T measured from t = 0, after which the splash is heard is given by ?

**Asked By: PREM**

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**Joshi sir comment**

$Byh=ut+\frac{1}{2}a{t}^{2},timeofdownwardjourneycanbecalculated\phantom{\rule{0ex}{0ex}}thenh/visthetimeforupwardjourneyofsound\phantom{\rule{0ex}{0ex}}nowsolve$

Sir pls solve

**Asked By: KANDUKURI ASHISH**

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**Joshi sir comment**

$Impulseis\overrightarrow{Ft}andPower=W/t\phantom{\rule{0ex}{0ex}}\overrightarrow{F}=m\overrightarrow{g}andW=mgh\phantom{\rule{0ex}{0ex}}Nowcalculatethetimeofmotion\phantom{\rule{0ex}{0ex}}andgettheanswer$

**Asked By: PRATEEK MOURYA**

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**Joshi sir comment**

$Accordingtogivencondition\phantom{\rule{0ex}{0ex}}\frac{d\theta}{dt}=\frac{v\mathrm{sin}60}{a-(v+v\mathrm{cos}60)t}\phantom{\rule{0ex}{0ex}}Nowsolveitfort\phantom{\rule{0ex}{0ex}}thendis\mathrm{tan}ce=vt$solve

inform for any difficulty

A long smooth cylindrical pipe of radius r is tilted at an angle alpha to the horizontal. A small body at point A is pushed upwards along the inner surface of the pipe so that the direction of its initial velocity forms an angle phi with generatrix AB. Determine the minimum initial velocity V at which the body starts moving upwards without being separated from the surface of the pipe

**Asked By: MAAZ**

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**Joshi sir comment**

At the front of the cylinder, see a tilted circle $Nowcomparethisproblemwithverticalcircularmotion.\phantom{\rule{0ex}{0ex}}Aisthetoppointofthatpath.Weknowthatthe\phantom{\rule{0ex}{0ex}}minimumvelocityattoppoint\mathrm{tan}gentialtopathof\phantom{\rule{0ex}{0ex}}verticalcircleshouldbe\sqrt{gr}.\phantom{\rule{0ex}{0ex}}Hereduetotiltofthiscircleeffectivegravity=g\mathrm{cos}\alpha \phantom{\rule{0ex}{0ex}}sov\mathrm{sin}\phi =\sqrt{g\mathrm{cos}\alpha r}\phantom{\rule{0ex}{0ex}}nowsolve\phantom{\rule{0ex}{0ex}}informforanyissue\phantom{\rule{0ex}{0ex}}waitforvideo$

**Asked By: KANDUKURI ASHISH**

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**Joshi sir comment**

According to first given condition velocity component of rain = u

let vertical component of rain = v

According to second condition

velocity of rain relative to man

$=(nu-u)(-\hat{i})+v(-\hat{j})$$heredirectionofmotionofmanis\hat{i}and\phantom{\rule{0ex}{0ex}}verticallyupwarddirectionisconsideredas\hat{j}\phantom{\rule{0ex}{0ex}}Nowdrawtriangleforthisvectorandgetcot\theta \phantom{\rule{0ex}{0ex}}solveit\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}informforanyissue$

A cylindrical pipe of radius r is rolling towards a frog sitting on the horizontal ground. Center of the pipe is moving with velocity v. To save itself ,the frog jumps off and passes over the pipe touching it only at top. Denoting air time of frog by T, horizontal range of the jump by R and acceleration due to gravity by g. Find T and R.

**Asked By: MUDIT**

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**Joshi sir comment**

$Accordingtogivencondition2r={{u}_{y}}^{2}/2g,\phantom{\rule{0ex}{0ex}}T=2{u}_{y}/g=2\sqrt{4gr}/g=4\sqrt{r/g}\phantom{\rule{0ex}{0ex}}Attoppointduetotouch,vecocityofthefrogbe{u}_{x}-2v.\phantom{\rule{0ex}{0ex}}R=[{u}_{x}{u}_{y}/g]+[{u}_{x}-2v){u}_{y}/g]=[{u}_{y}/g\left]\right(2{u}_{x}-2v)\phantom{\rule{0ex}{0ex}}=[2{u}_{y}/g\left]\right({u}_{x}-v)=T({u}_{x}-v)\phantom{\rule{0ex}{0ex}}Nowminimumpossiblevalueofvelocityoffrogattopto\phantom{\rule{0ex}{0ex}}crossthecylinderis\sqrt{gr}(Bytheconceptofverticalcircle)\phantom{\rule{0ex}{0ex}}nowsolvecarefullywithouterror$

A small ball is thrown from foot of a wall with minimum possible velocity to hit a bulb B on the ground at a distance L away from the wall.Find the expression for height h of shadow of the ball on the wall as a function of time t.Acceleration due to gravity is g.

**Asked By: MUDIT**

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**Joshi sir comment**

While making this paper, I stayed up till late at night working with candle light. I had two

candles of equal length L lit and placed at the same time, as shown in the figure. Immediately

thereafter I noticed that the shadow of the first candle on the left wall is not changing in

length, and the shadow of the second candle on the right wall is being shortened at a speed v.

(Take d1

= 10 cm, d2

= 20 cm, d3

= 30 cm, v = 1 mm/s, L = 20 cm.)

**Asked By: AAYUSH MALAVIYA**

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**Joshi sir comment**

Problem is interesting yet not complete. Complete it to get answer

thanks