Diyadiya
  • Diyadiya
Question:-
Physics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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Diyadiya
  • Diyadiya
this is the equation right ?\[s=ut+ \frac{1}{2} gt^2\]
Diyadiya
  • Diyadiya
@imranmeah91
anonymous
  • anonymous
you are missing , initial height

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anonymous
  • anonymous
\[x=x_0+v_0 t+1/2 a t^2\]
Diyadiya
  • Diyadiya
i forgot all these things :S
anonymous
  • anonymous
it is important , 4 kinematic eqauation http://www.physicsclassroom.com/class/1dkin/u1l6a.cfm
Diyadiya
  • Diyadiya
where did that initial height come from?
anonymous
  • anonymous
if you throw a projectile from top of a building , you will have to factor in the height of the building
Diyadiya
  • Diyadiya
i thought it was just \[s=ut+ \frac{1}{2} gt^2\]
anonymous
  • anonymous
only if the intial height is zero
Diyadiya
  • Diyadiya
oh
Diyadiya
  • Diyadiya
so stone falls from the top of the tower in 8s
anonymous
  • anonymous
yes
anonymous
  • anonymous
but initially it is at the Top of the tower
Diyadiya
  • Diyadiya
so what's x_o here ?
anonymous
  • anonymous
height of the tower which we just call as h
Diyadiya
  • Diyadiya
ok and what is x here?
anonymous
  • anonymous
well , after 8 second , it is on the ground which mean height is equal to 0
Diyadiya
  • Diyadiya
ok so x is final height ?
anonymous
  • anonymous
yes
Diyadiya
  • Diyadiya
and u is 0?
anonymous
  • anonymous
yes, intially there is no velocity , like the instant when you let go
Diyadiya
  • Diyadiya
yeah so 0=h+1/2*9.8*64 ?
anonymous
  • anonymous
yes
Diyadiya
  • Diyadiya
can we take g as 10 ?
anonymous
  • anonymous
well, we will lose precision
Diyadiya
  • Diyadiya
ok
Diyadiya
  • Diyadiya
so h=-313.6
anonymous
  • anonymous
should be positive
anonymous
  • anonymous
g=-9.8
Diyadiya
  • Diyadiya
Oh ok
anonymous
  • anonymous
0=h+1/2*(-9.8)*64
Diyadiya
  • Diyadiya
ok 313.6
anonymous
  • anonymous
ok , use that in other equation
Diyadiya
  • Diyadiya
ok
Diyadiya
  • Diyadiya
Well ,h-h/4=h+0*t+ 1/2 * 9.8* (t^2) what is that?
anonymous
  • anonymous
well, final height is first quarter height of tower
anonymous
  • anonymous
so h- h/4
Diyadiya
  • Diyadiya
oh ok
apoorvk
  • apoorvk
i guess the problem is solved, right?
Diyadiya
  • Diyadiya
No
Diyadiya
  • Diyadiya
i didn't get it
anonymous
  • anonymous
ok what part?
Diyadiya
  • Diyadiya
h-h_o/4=1/2*-9.8*t^2 ?
apoorvk
  • 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
  • Diyadiya
I haven't done all these since long time ,so i completely forgot yeah i'll post
anonymous
  • 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
  • Diyadiya
ok h is zero right?
anonymous
  • anonymous
no, h as you found out was 313
Diyadiya
  • Diyadiya
OHHHHHHHHHH YA
Diyadiya
  • Diyadiya
ok so now i have to find t ?
anonymous
  • anonymous
yes
Diyadiya
  • 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
  • Diyadiya
Okay 4s :D
anonymous
  • anonymous
good job,
Diyadiya
  • Diyadiya
THANK YOU SO MUCH!
Diyadiya
  • Diyadiya
whats the angle b/w instantaneous displacement and acceleration during the related motion?
apoorvk
  • 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
  • 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
  • Diyadiya
Thankyouu i'll try this as well but when is g +ve and -ve?
apoorvk
  • 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
  • Diyadiya
Ok yeah got it :D thanks

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