Two stone are thrown form the same point at the same time, vertically upwards with speed 30ms-1, and the other vertically downwards at 30ms-1. Find how far apart the stone are after 3 seconds.
Thanks a lot
Mechanics question?
2021-01-11 11:47 am
回答 (2)
2021-01-11 12:21 pm
✔ 最佳答案
Take g = 9.8 m/s²h = ut - (1/2) gt²
For the first stone:
h = (30)(3) - (1/2)(9.8)(3)²
Displacement, h = 45.9 m
For the second stone:
h = (-30)(3) - (1/2)(9.8)(3)²
Displacement, h = -134.1 m
Distance apart of the two stones after 3 seconds
= 45.9 - (-134.1)
= 180 m
2021-01-11 4:08 pm
The solution becomes still easy if you assume the first stone moving down wards ( instead of upwards) with an initial velocity of (-)ve 30 m/s and g +ve. Herte you wont need to put the value of g since it gets cancelled out.
收錄日期: 2021-04-24 08:14:09
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20210111034713AAgTiEt