1. ## stone

a stone is dropped from the top of a building 30 m high. a second stone is dropped from half way up the same building.. find the time that should elapse between the release of the 2 stones if they are if they are to reach the ground at the same time

stone released from top; u=0 s=30 a=g t=t+x stone half way; u=0 s=15 a=g t=t
x stands for extra time the top stone was relased from. when i use s=ut+.5t^2 i cant work it out. equations could be wrong

2. ## Re: stone Originally Posted by markosheehan a stone is dropped from the top of a building 30 m high. a second stone is dropped from half way up the same building.. find the time that should elapse between the release of the 2 stones if they are if they are to reach the ground at the same time

stone released from top; u=0 s=30 a=g t=t+x stone half way; u=0 s=15 a=g t=t x stands for extra time the top stone was relased from. when i use s=ut+.5t^2 i cant work it out. equations could be wrong
Find the time for the first stone to reach the ground. Then do the same for the second. Then subtract the times.

3. ## Re: stone Originally Posted by markosheehan a stone is dropped from the top of a building 30 m high. a second stone is dropped from half way up the same building.. find the time that should elapse between the release of the 2 stones if they are if they are to reach the ground at the same time

stone released from top; u=0 s=30 a=g t=t+x stone half way; u=0 s=15 a=g t=t x stands for extra time the top stone was relased from. when i use s=ut+.5t^2 i cant work it out. equations could be wrong
u= 0 for both as you say but you left "g" out of your equation. You should be using s= .5gt^2.

30= .5g(t+ x)^2= .5gt^2+ gtx+ .5gx^2 and 15= .5gt^2. Subtracting, gtx+ .5gx^2= 15. First solve .5gt^2= 15 to find t, then, using that value of t, solve .5gx^2+ gtx- 15= 0 for x.

(That is effectively the same as Debsta suggested.)