A ball is kicked off a pier & lands in the sea. h(t)=2/7(t+3)(7-t) h=height, t = secs. When does the ball land in the sea?
回答 (4)
2016-06-22 6:03 am
✔ 最佳答案
h = (2/7)(t + 3)(7 - t)When h = 0 :
(2/7)(t + 3)(7 - t) = 0
(t + 3)(7 - t) = 0
t = -3 (rejected) or t = 7
The ball lands in the sea in 7 seconds.
2016-06-22 12:48 pm
Max height = dh/dt=0
2/7 ((7-t) -(t+3))=0
This gives
4=2t,
t=2, This much time to come down. That will 4 seconds.
2/7 ((7-t) -(t+3))=0
This gives
4=2t,
t=2, This much time to come down. That will 4 seconds.
2016-06-22 6:17 am
I'll assume that the height is measured above sea level not the level of the pier. If so we simply want the time 't' when the height is zero:
0 = 2/7(t+3)(7-t). Now just look at this equation; the only positive value for t that evaluates correctly is t = 7.
0 = 2/7(t+3)(7-t). Now just look at this equation; the only positive value for t that evaluates correctly is t = 7.
2016-06-22 5:22 am
what type of ball? different balls travel at different velocities.
收錄日期: 2021-04-18 15:14:01
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160621212004AAXAcK5