IIT JEE , AIEEE Forum - Physics , Mechanics

Avinash_Bhat

KINEMATICS : 1 DIMENSIONAL

body dropped from height ' h ', it reaches the ground in time ' t ' such that final velocity is ' u '. n

R

At height (h - h/n) from the ground : the velocity is u /

n

: the time taken is t /

n

body dropped from height ' h ' while at the sametime another body vertically projected upwards with a velocity ' u '. these meet after a time of (h / u) at a distance of (g / 2)(h / u)²from the top.

' n ' balls thrown every second such that, one ball thrown when the previous one reaches its maximum height. Now this max. height = (g / 2n² )

Juggler keeps n balls going with one hand, so that (n-1) balls in air and 1 in hand. Each ball rises to height 'x' so that time for each ball to stay in his hand is : 2

(2x/g) / (n-1)

Rucha_Vaidya

Hey, this is a Questions discussion panel and u r writing article. You must write ur articles in community shelf.

Avinash_Bhat

ball vertically thrown up, passes through the same point at times, t1 and t2.

Then : t1 + t2 = 2u / g = total time the ball remains in air ( T ).

: t1*t2 = T²

body dropped from a height and the time taken to cover successive 1 m distances in the ratio 1 : (

2 -

1) : (

3 -

2) : (

4 -

3) : (

5 -

4) : .......

body starts from rest and constantly accelerates at 'a' m/s² for time t1 and then retards at 'r' m/s² for time t2, coming to rest. Let t1 + t2 = T Maximum velocity attained = at1 = rt2 = ( arT) / (a + r). Total diatance covered = (arT² ) / 2(a + r)

What do you say ??

Avinash_Bhat

Hey Rucha, I see that you are a newcomer, still I appreciate your keen outlook. Here take this rate ....

We usually discuss these too Rucha........After all ...this is the "Discuss with community" panel and we are here to do something, to become IITians !! ......

I just want to share my knowledge so that I could help everyone, especially our friends VMC,deep and others who have asked me to write these...

Avinash_Bhat

body dropped from the top of a tower and after 'n' seconds another stone is thrown downwards with velocity ' v '. They can meet after a time of :

{ (gn/2 - v) / (gn - v) }n

bus accelerates at a constant rate 'a' from rest. man standing at a distance of 'd' away from the bus should run at a minimum velocity of v =

(2ad) to get to the bus.

body uniformily accelerares from rest at a rate a1 for time t1 and then at a2 for t2, at a3 for t3 .........

avg. acceleration is : (a1t1 + a2t2 + a3t3 + ....) / (t1 + t2 + t3 +....)

I will be back.....

joyfrancis

it is a lot easier to solve the problem than to remeber so many formulaes. Atleast for me.
I still appreciate your help though ...

VMC

hey avinash u have already posted this article 1 month before so why to repeat it???pls give us new formulas!! pls!!!!!!

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