A tennis ball on Mars, where the acceleration due to
gravity is 0.379g and air resistance is negligible, is hit
directly upward and returns to the same level 8.5s later.
a) How high above its original point did the ball go?
b) How fast was it moving just after being hit?
tennis ball on Mars
2013-05-29 12:49 am
回答 (1)
2013-05-29 2:23 am
✔ 最佳答案
(a) Use equation: s = vt - (1/2)at^2with v = 0 m/s, a = -0.379g, t = 8.5/2 s = 4.25 s, s =?
hence, s = -(1/2).(-0.379g).(4.25^2) m = 34.23 m
[Take g = 10 m/s^2]
(b) Use v = u + at
with v = 0 m/s, a = -0.379g , t = 4.25 s, u =?
0 = u - 0.379g x 4.25
u = 16.1 m/s
收錄日期: 2021-04-29 17:50:26
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20130528000051KK00157