ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: luken93 on October 21, 2010, 07:55:38 pm
-
Two Stones are thrown from the top of a building at the same instant. The first stone is thrown vertically upward and reaches the ground in 4 seconds.
The second stone is thrown vertically downward with the same initial speed as the first stone and reaches the ground in 2 seconds.
(Use 9.8m/s2 as acceleration)
1) Find the height of the building and the initial speed of each stone.
2) Find the maximum distance between the two stones during their motion
3) A third stone is dropped from the top of the building and reaches the ground at the same instant as the first stone. How long after the first two stones were thrown was the third dropped
I got 1) and 3), but it was a fairly long process, so wondering if there is anyway to shorten it as well as a method to find out 2)
Thanks
-
I think for 2) the maximum distance will be when the second stone hits the ground.
This is because stone 2 is always travelling faster downwards than stone 1 (until it hits the ground of course) and hence the distance between will always be increasing up until that point.
ps. that's assuming the first stone has already reached it's apex and is travelling downwards by this point. If not, then the max distance is simply the max height of the first stone.
-
I think for 2) the maximum distance will be when the second stone hits the ground.
This is because stone 2 is always travelling faster downwards than stone 1 (until it hits the ground of course) and hence the distance between will always be increasing up until that point.
ps. that's assuming the first stone has already reached it's apex and is travelling downwards by this point. If not, then the max distance is simply the max height of the first stone.
hmmm thats quite a good way of doing it, just find the time when the first hits the ground, as that will be the greatest distance between...