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- Diyadiya

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- Diyadiya

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

- Diyadiya

@imranmeah91

- anonymous

you are missing , initial height

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## More answers

- anonymous

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

- Diyadiya

i forgot all these things :S

- anonymous

it is important , 4 kinematic eqauation
http://www.physicsclassroom.com/class/1dkin/u1l6a.cfm

- Diyadiya

where did that initial height come from?

- anonymous

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

- Diyadiya

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

- anonymous

only if the intial height is zero

- Diyadiya

oh

- Diyadiya

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

- anonymous

yes

- anonymous

but initially it is at the Top of the tower

- Diyadiya

so what's x_o here ?

- anonymous

height of the tower which we just call as h

- Diyadiya

ok and what is x here?

- anonymous

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

- Diyadiya

ok so x is final height ?

- anonymous

yes

- Diyadiya

and u is 0?

- anonymous

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

- Diyadiya

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

- anonymous

yes

- Diyadiya

can we take g as 10 ?

- anonymous

well, we will lose precision

- Diyadiya

ok

- Diyadiya

so
h=-313.6

- anonymous

should be positive

- anonymous

g=-9.8

- Diyadiya

Oh ok

- anonymous

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

- Diyadiya

ok 313.6

- anonymous

ok , use that in other equation

- Diyadiya

ok

- Diyadiya

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

- anonymous

well, final height is first quarter height of tower

- anonymous

so h- h/4

- Diyadiya

oh ok

- apoorvk

i guess the problem is solved, right?

- Diyadiya

No

- Diyadiya

i didn't get it

- anonymous

ok what part?

- Diyadiya

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

- apoorvk

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?

- Diyadiya

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

- anonymous

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

- Diyadiya

ok h is zero right?

- anonymous

no, h as you found out was 313

- Diyadiya

OHHHHHHHHHH YA

- Diyadiya

ok so now i have to find t ?

- anonymous

yes

- Diyadiya

@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

- Diyadiya

Okay 4s :D

- anonymous

good job,

- Diyadiya

THANK YOU SO MUCH!

- Diyadiya

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

- apoorvk

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!!

- apoorvk

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

- Diyadiya

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

- apoorvk

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.

- Diyadiya

Ok yeah got it :D thanks

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