h(t)=-16t^2+80t. whats the maximum height that the rocket will reach?
2016-11-15 7:57 pm
I'm not sure how to solve this so if you could tell me the formula, that'd be great!
回答 (4)
2016-11-15 8:03 pm
This is a parabola, so put it in vertex form:
h(t) = −16t² + 80t
h(t) = −16 (t²−5t+25/4) + 16(25/4)
h(t) = −16 (t−5/2)² + 100
Maximum height = 100 ft, when t = 2.5 sec
2016-11-15 9:44 pm
v =-32t +80=, v=0
0=-32t+80
32t=80 then t 80/32=5/2 (max height at t=5/2)
h(5/2)=-16(5/2)^2+80(5/2)=-16(25)/4 +400/2
=-4(25)+200=-100+200=100
0=-32t+80
32t=80 then t 80/32=5/2 (max height at t=5/2)
h(5/2)=-16(5/2)^2+80(5/2)=-16(25)/4 +400/2
=-4(25)+200=-100+200=100
2016-11-15 8:11 pm
You need to find the vertex, either by putting the equation into vertex form, or by finding the point where the first derivative equals 0.
h(t) = -16t² + 80t
h'(t) = -32t + 80 = 0
32t = 80
t = 80/32 = 2.5
max height = h(2.5) = 100 ft
h(t) = -16t² + 80t
h'(t) = -32t + 80 = 0
32t = 80
t = 80/32 = 2.5
max height = h(2.5) = 100 ft
2016-11-15 8:00 pm
H(t) = at^2 + bt + c
the maximum height = H(-b/(2a))
the maximum height = H(-b/(2a))
收錄日期: 2021-05-01 21:18:04
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20161115115708AAAWYq6