IIT JEE , AIEEE Forum - Physics , Mechanics

Avinash_Bhat

'KINEMATICS'

1 DIMENSIONAL MOTION

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

R

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

body dropped from a height and the time taken to cover successive 1 m distances are in the ratio is 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)

chimanshu_007

gr8 work
plz rply so that i can rate u

chimanshu_007

plz see these too
http://www.goiit.com/posts/list/7067.htm

chimanshu_007

http://www.goiit.com/posts/list/6412.htm

Avinash_Bhat

body accelerates at a constant rate 'a' for sometime from rest, and then with a constant velocity ' v ' moves for sometime, and retards at a rate 'r' coming to rest covering a distance ' x ',

then the total time taken = (x/v) + v/2(1/a + 1/r)

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

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

A bus accelerates at a constant rate 'a' from rest. A 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.

Avinash_Bhat

Hey Himanshu, where are these posts(i am a newcomer, yaar) ?

Avinash_Bhat

Nothing, nothing I got it....

chimanshu_007

ok
so saw them?

Avinash_Bhat

Yaa. Excellent work, dude.

Avinash_Bhat

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

Then the average acceleration is : (a1t1 + a2t2 + a3t3 + ....) / (t1 + t2 + t3 +....)

PROJECTILES (2 - DIMENSIONAL):

for a vertical projectile : y = usin

- gt² / 2

= xtan

- gx² / 2u²cos²

H = t²(g / 8)

body dropped from the top of a building of height 'h'. Another body is thrown upwards from the ground level with a velocity 'u' making

with the ground.

OR

body thrown horixontally downwards with a velocity ' v ' from a height 'h' while another body is thrown upwards from the ground level with a velocity 'u' making

with the ground.

In both cases : they can meet after a time of : (h / usin

)

Avinash_Bhat

Body A thrown upwards with a velocity 'u' making

1 with the ground collides with another body B, which is thrown with a velocity ' v ', from the other side, making an angle

2 with the ground. If they meet after time 't', at a horizontal distance of 'a' and 'b' along x-axis from bodies A and B respectively, then :

t = x / ucos

1 = y / vcos

2

t1 is the time of flight when body is projected at

with the horixontal and t2 is the time of flight when the body is projected at (90 -

) then :

=> t1 * t2 = 2R / g

=> t1 / t2 = tan

=> The heights H1 and H2 during t1 and t2 relates R as : R = 4

(H1 * H2)

=> H1 / H2 = tan²

Avinash_Bhat

Projectile projected with velocity 'u' making angle# with the plane from the bottom A of an inclined plane(making angle@ with the ground) so that it reaches B. (Let C be a point on the ground such that AC perp. to BC)

Time taken,t : 2u sin# / g cos@

AC : u cos(# + @) t

: 2u^2 sin# cos(# + @) / g cos@

AB : AC / gcos@

Avinash_Bhat

projectile thrown horizontally down with a velocity 'u' from the top A of an inclined plane making an angle$ with the plane so that it reaches B on the inclined plane. (hope you got that inclined plane makes angle$ with the ground)

Time taken : 2u tan$ / g

AB : 2u^2 tan$ / gcos$

Vertical distance covered : AB * cos$

projctile projected from the bottom of an inclined plane with a velocity 'u' making angle$ with the plane while another one is projected downwards from the top of the plane with velocity ' v ' making angle@ with the plane.

They meet if : usin$ = vsin@

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