# 95 - Kinematics Questions Answers

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

Two boys enter a running escalator at the ground floor of a shopping mall . The first boy repeatedly follows a cycle of p1=1 step up and then q1=2 steps down whereas the second boy repeatedly follows a cycle of p2=2 steps up and then q2=1 step down. Both of them move relative to escalator with a speed v=50cm/s . If the boys take t1=250s and t2=50s respectively to reach the first floor in complete number of cycles, how fast is the escalator running ?

**Asked By: JATIN KUMAR**

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**Joshi sir comment**

$letuandvarethespeedsofboyandescalator\phantom{\rule{0ex}{0ex}}accordingtogivencondition\phantom{\rule{0ex}{0ex}}\left[1\right(u+v)T-2(u-v\left)T\right]{n}_{1}=\left[2\right(u+v)T-1(u-v\left)T\right]{n}_{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \left[1\right(u+v)-2(u-v\left)\right]T{n}_{1}=\left[2\right(u+v)-1(u-v\left)\right]T{n}_{2}\phantom{\rule{0ex}{0ex}}GiventhatT{n}_{1}=250andT{n}_{2}=50\phantom{\rule{0ex}{0ex}}\Rightarrow \left[1\right(u+v)-2(u-v\left)\right]250=\left[2\right(u+v)-1(u-v\left)\right]50\phantom{\rule{0ex}{0ex}}nowsolveitforv,uisgiven$

A particle projected from the ground passes two points , which are at height 12m and 18m above the ground and a distance d=10m apart . What could be the minimum speed of projection. Acceleration due to gravity is g=10m/s^2 .

**Asked By: JATIN KUMAR**

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**Joshi sir comment**

the min no. of non coplanar and non zero vectors of unequal magnitude which can give zero resultant is _________??

**Asked By: SARIKA**

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**Joshi sir comment**

four

let three are i, j, k then fourth will be -(i+j+k)

if y= 50 / [(x-5)^{2} +5]^{2} then maximum value of y will be??

**Asked By: SARIKA**

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**Joshi sir comment**

let z = [(x-5)^{2} +5]

then dz/dx = 2(x-5)

compare it to 0 we get x = 5

so for x = 5, z will be minimum and so y will be max for that x

put x= 5 and get answer

A particle has initial velocity of y = 3i+ 4j and a constant force F = 4i -3j acts on it. Then what will be the path followed by the particle?

**Asked By: VIKRAM**

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**Joshi sir comment**

acceleration is constant and initial velocity is perpendicular to it because dot product of two given vectors is 0 so this is the case equivalent to a particle thriwn from the top of a building horizontally under gravity so path will be parabolic

if you want the equation of path then reply