Question

Question:-

Answer 1

this is the equation right ?\[s=ut+ \frac{1}{2} gt^2\]

Answer 2

@imranmeah91

Answer 3

you are missing , initial height

Answer 4

\[x=x_0+v_0 t+1/2 a t^2\]

Answer 5

i forgot all these things :S

Answer 6

it is important , 4 kinematic eqauation

http://www.physicsclassroom.com/class/1dkin/u1l6a.cfm

Answer 7

where did that initial height come from?

Answer 8

if you throw a projectile from top of a building , you will have to factor in the height of the building

Answer 9

i thought it was just \[s=ut+ \frac{1}{2} gt^2\]

Answer 10

only if the intial height is zero

Answer 11

oh

Answer 12

so stone falls from the top of the tower in 8s

Answer 13

yes

Answer 14

but initially it is at the Top of the tower

Answer 15

so what's x_o here ?

Answer 16

height of the tower which we just call as h

Answer 17

ok and what is x here?

Answer 18

well , after 8 second , it is on the ground which mean height is equal to 0

Answer 19

ok so x is final height ?

Answer 20

yes

Answer 21

and u is 0?

Answer 22

yes, intially there is no velocity , like the instant when you let go

Answer 23

yeah
so
0=h+1/2*9.8*64 ?

Answer 24

yes

Answer 25

can we take g as 10 ?

Answer 26

well, we will lose precision

Answer 27

ok

Answer 28

so
h=-313.6

Answer 29

should be positive

Answer 30

g=-9.8

Answer 31

Oh ok

Answer 32

0=h+1/2*(-9.8)*64

Answer 33

ok 313.6

Answer 34

ok , use that in other equation

Answer 35

ok

Answer 36

Well ,h-h/4=h+0*t+ 1/2 * 9.8* (t^2)
what is that?

Answer 37

well, final height is first quarter height of tower

Answer 38

so h- h/4

Answer 39

oh ok

Answer 40

i guess the problem is solved, right?

Answer 41

No

Answer 42

i didn't get it

Answer 43

ok what part?

Answer 44

h-h_o/4=1/2*-9.8*t^2 ?

Answer 45

i guess you need a bit of revision. no worries there.
can you post an exact problem @diyadiya so that we can show how the things are working, and clear out the cloud?

Answer 46

I haven't done all these since long time ,so i completely forgot
yeah i'll post

Answer 47

actually , it is

final height= initial hieght+ vo * t + 1/2 a t^2

h-h/4=h+ v0t + 1/2 a t^2

two h cancels, vo =0

-h/4= 1/2 (9.8) t^2

Answer 48

ok h is zero right?

Answer 49

no, h as you found out was 313

Answer 50

OHHHHHHHHHH YA

Answer 51

ok so now i have to find t ?

Answer 52

yes

Answer 53

@apoorvk A stone falls from the top of the tower in 8s. How much time it will take to cover the 1st quarter of the distance starting from top

Answer 54

Okay 4s :D

Answer 55

good job,

Answer 56

THANK YOU SO MUCH!

Answer 57

whats the angle b/w instantaneous displacement and acceleration during the related motion?

Answer 58

okay. so we dissect this question now. here's how. since i dont know what 1/4th of the height is, let me first find the entire height. and then we 'll try to solve for the time taken to fall that distance.
let the entire height be 'h',
intial velocity= u=0 (since its been 'dropped')
time taken to cover = t= 8 sec.
and acceleration = +g ('+' since the gravity here wil have an accelerating effect on the ball's speed with time)
so, using second equation of motion,
\[h = ut + (1/2) g t^2\]

substitute u=0, g=10 or 9.8 whichever, t= 8

you get the entire ht 'h' which is something like 320 m (if i am right)

now one quarter of that height, that is 'h/4' is what? 80 metres yes!!

so once again we remember newton and his second equation
\[h/4 = ut + (1/2) g t^2\]

you know what h/4 is, you know u=0. and you know g=10

so you have a quadratic in 't'.

so, what next?? solve it for 't'!! get time required. (and if you are just wondering what to do with negative values of 'time', just drop 'em!! time can't be negative in the context we 're talking!)
what answer do you get? was 4 sec what you got there?


hope that brought back some good ol' memories!!

Answer 59

just a point, in the secong part take the time as " t' " just to avoid confusion with total time, which was 't=8'.

Answer 60

Thankyouu i'll try this as well
but when is g +ve and -ve?

Answer 61

see. this is a very confusing thing for some (though it actually isn't at all), and different people use it in a different way. here's how i use it (and you can too if it suits you):

when i see that the object is under a free-fall, i take g as '+' because if g is positive, it means the body is under positive acceleration, and its speed increases as it falls (that is what happens, kya?).

now if the body is being thrown up, it stops at some height above, so basically you get a feel it's undergoing retardation, or negative acceleration. so there i take g as "-ve", so that the retardation can be effected in the equation.

Answer 62

Ok yeah got it :D thanks