In a charity show, a car 'flies' from the roof of one building to another building which is 10 m lower.
If the initial velocity of the car is 25 m s^-1 in the horizontal direction, what is the maximum distance between the buildings below which the car can land safely?
Physics: force and motion
2012-02-11 1:31 am
回答 (1)
2012-02-11 1:42 am
✔ 最佳答案
First of all, the time of flight allowed is to be found on the basis of the vertical displacement:s = ut + at2/2 with u = 0 (initial vertical velocity = 0), a = 10 and s = 10:
10 = 5t2
t = 1.41s
So the max. horizontal displacement is:
1.41 x 25 = 35.4 m
which is also the max. distance between the buildings
參考: 原創答案
收錄日期: 2021-04-13 18:31:33
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20120210000051KK00520