Acceleration due to gravity pls HELP

[mickarose]

hi there...
am tearing my hair out over a question to do with acceleration due to gravity...

a stone is dropped from rest, from a height of 20m; 0.6 seconds later a marble is thrown with a velocity of 8m/s from the same height. when does the marble overtake the stone..?(assuming gravity to be 10m/s~2.)

ive spent so much time on this but am really confused about the time factor..
i know that it overtakes when the displacement is the same so when
ut+1/2at^2=ut+1/2at^2 but I am confused that one has a head start of .6 sec...

ive tryed working out that the stone has a 1.8m displacement after .6 seconds and added that to the formula before making both sides equal but what i end up with is


(stone)1.8m + 5t^2 =(marble) 8t + 5t^2

dont know what I am doing wrong... anyone that can help out id be greatly appriciative...

 

[h_k331]

I'll take a shot at it for you. I'm just learning this stuff too so I hope I don't misguide you.

Position as a function of time can be described by xf = xi + vi*t + 0.5*a*t^2 where xf is final position, xi is initial position, vi is initial velocity, a is acceleration, and t is time. What you are looking for is the time when the final position (xf) of the stone is equal to the final position of the marble. To find this you need to set up the above equation for xf twice, once for the stone and once for the marble. After you set up these two separate equations you can set them equal to each other and solve them for time (t).

Treat the ground as your origin, so for the stone:
xi = 20.0 m
vi = 0 m/s
a = -10.0 m/s^2
t = t + 0.6 sec

For the marble:
xi = 20.0 m
vi = 8.0 m/s
a = -10.0 m/s^2
t = t

Your answer will be in seconds after the marble is thrown. For seconds after the stone is dropped simply add 0.6 seconds to your final answer. Someone yell at me if I'm giving bad advice.

hk

 

[mickarose]

hmm...still stuck

ok, thanks muchly for the reply...

will take me a while to ponder it...

not seen it done like that b4...

my head hurts..lol

 

[VietDao29]

There is 1 mistake in your work. This line is wrong:

(stone)1.8m + 5t^2 =(marble) 8t + 5t^2

This is wrong because after 0.6 second, the stone has the speed of : 10 * 0.6 = 6 m / s (not 0 m / s).
So that line should read: (stone)1.8m + 6t + 5t^2 =(marble) 8t + 5t^2
Solve that equation and you will have the amount of time needed for the marble to catch up with the stone after the marble was thrown down.
Viet Dao,

 

[mickarose]

awsome, thanks very much for your help...
knew i was close but really was bugging me i couldn't get it...
get the right ans using you solution vietdao...
cheers...