114 - Kinematics Questions Answers

Physics >> Mechanics part 1 >> Kinematics IIT JEE

A projectile is fired with velocity u at entrance A of a horizontal tunnel of length L and height H. The minimum value of u for which projectile will reach the other end of the tunnel without touching the top of the tunnel is

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Physics >> Mechanics part 1 >> Kinematics IIT JEE

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Physics >> Mechanics part 1 >> Kinematics IIT JEE

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 ?

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$B y h = u t + \frac{1}{2} a t^{2}, t i m e o f d o w n w a r d j o u r n e y c a n b e c a l c u l a t e d t h e n h / v i s t h e t i m e f o r u p w a r d j o u r n e y o f s o u n d n o w s o l v e$

Physics >> Mechanics part 1 >> Kinematics IIT JEE

Sir can you please solve this question??? Thank you very much sir

Asked By: GOLU

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Physics >> Mechanics part 1 >> Kinematics IIT JEE

Asked By: SYED SAHIB AKTAR

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Physics >> Mechanics part 1 >> Kinematics IIT JEE

Sir pls solve

Asked By: KANDUKURI ASHISH

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$I m p u l s e i s \vec{F t} a n d P o w e r = W / t \vec{F} = m \vec{g} a n d W = m g h N o w c a l c u l a t e t h e t i m e o f m o t i o n a n d g e t t h e a n s w e r$

Physics >> Mechanics part 1 >> Kinematics IIT JEE

Sir please make a video explaining this problem solution as i love your explanation

Asked By: PRATEEK MOURYA

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$A c c o r d i n g t o g i v e n c o n d i t i o n \frac{d θ}{d t} = \frac{v \sin 60}{a - (v + v \cos 60) t} N o w s o l v e i t f o r t t h e n d i s \tan c e = v t$ solve

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Physics >> Mechanics part 1 >> Kinematics IIT JEE

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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At the front of the cylinder, see a tilted circle $N o w c o m p a r e t h i s p r o b l e m w i t h v e r t i c a l c i r c u l a r m o t i o n . A i s t h e t o p p o i n t o f t h a t p a t h . W e k n o w t h a t t h e m i n i m u m v e l o c i t y a t t o p p o i n t \tan g e n t i a l t o p a t h o f v e r t i c a l c i r c l e s h o u l d b e \sqrt{g r} . H e r e d u e t o t i l t o f t h i s c i r c l e e f f e c t i v e g r a v i t y = g \cos α s o v \sin φ = \sqrt{g \cos α r} n o w s o l v e i n f o r m f o r a n y i s s u e w a i t f o r v i d e o$

Physics >> Mechanics part 1 >> Kinematics IIT JEE

Asked By: KANDUKURI ASHISH

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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

$= (n u - u) (- \hat{i}) + v (- \hat{j})$ $h e r e d i r e c t i o n o f m o t i o n o f m a n i s \hat{i} a n d v e r t i c a l l y u p w a r d d i r e c t i o n i s c o n s i d e r e d a s \hat{j} N o w d r a w t r i a n g l e f o r t h i s v e c t o r a n d g e t c o t θ s o l v e i t i n f o r m f o r a n y i s s u e$

Physics >> Mechanics part 1 >> Kinematics IIT JEE

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

$A c c o r d i n g t o g i v e n c o n d i t i o n 2 r = {u_{y}}^{2} / 2 g, T = 2 u_{y} / g = 2 \sqrt{4 g r} / g = 4 \sqrt{r / g} A t t o p p o i n t d u e t o t o u c h, v e c o c i t y o f t h e f r o g b e u_{x} - 2 v . R = [u_{x} u_{y} / g] + [u_{x} - 2 v) u_{y} / g] = [u_{y} / g] (2 u_{x} - 2 v) = [2 u_{y} / g] (u_{x} - v) = T (u_{x} - v) N o w m i n i m u m p o s s i b l e v a l u e o f v e l o c i t y o f f r o g a t t o p t o c r o s s t h e c y l i n d e r i s \sqrt{g r} (B y t h e c o n c e p t o f v e r t i c a l c i r c l e) n o w s o l v e c a r e f u l l y w i t h o u t e r r o r$

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