A Physics Question !
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Use the freaking formulas in this webby lar. http://www.stvincent.ac.uk/Resources/Physics/Speed/speed/formulae.html
s = ut + ½ at^2
then, S = 0.5 * 0.1 * 20* 20.
Hence, Distance travelled.
since the stone need to travel the same distance,
0.5 * 0.1 * 20* 20 = 0.5 * 10 * t^2
Resolve yourself . -
Its 9.80665E +00m/s^2, don't argue with me.
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The boy like Mac or KFC?
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a) 20mOriginally posted by _bluey_:Hey all , just wanna ask u guys a qn from my exam , kinda forget the whole part , so just a rough summary...
A boy is on a crane , and the crane accelerate from ground at 0.1m per second square. The crane accelerate for 20seconds. The boy dropped a stone down.
a)Find how high the boy is above the ground.
b)Find the time taken for the stone to drop to the ground.
Anyone knows?
b) take g=9.81, time taken=2.019s
not sure abt the answer cos have some doubts abt the qn...and so long nvr do physics liao
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Anyone tried the formula F = uck???
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Originally posted by mochou:Anyone tried the formula F = uck???
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Originally posted by ctstalin:a) 20m
b) take g=9.81, time taken=2.019s
not sure abt the answer cos have some doubts abt the qn...and so long nvr do physics liao good work for the part a) ! Can tell me how u solve for part b? 
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your one so chimOriginally posted by SexyChin:Use the freaking formulas in this webby lar. http://www.stvincent.ac.uk/Resources/Physics/Speed/speed/formulae.html
s = ut + ½ at^2
then, S = 0.5 * 0.1 * 20* 20.
Hence, Distance travelled.
since the stone need to travel the same distance,
0.5 * 0.1 * 20* 20 = 0.5 * 10 * t^2
Resolve yourself .
and where did 0.5 came from 
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0.5 came from 1/2.Originally posted by _bluey_:your one so chim and where did 0.5 came from -
use the formula s = ut +1/2at^2, since u=0, thereforeOriginally posted by _bluey_: good work for the part a) ! Can tell me how u solve for part b?
20 = 1/2(9.81).t^2
t= 2.019s
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OKAY ! sorta understood already , but why u=0 ? that's the only problem , i thought is 2ms or something? Since the crane is moving upwards =S.Originally posted by ctstalin:use the formula s = ut +1/2at^2, since u=0, therefore
20 = 1/2(9.81).t^2
t= 2.019s
Cannot differentiate between when is u=0 , and when is u = *more then 0 * for the cases when there something moving upwards or downwards leh. Eg , balloon , crane etc etc.
Thanks alot btw ! -
nice oneOriginally posted by ditzy:0.5 came from 1/2.