rate of changes
2013-10-03 2:30 am
1. a stone is thrown from the edge of a bridge that is 48 ft above the ground with an initial velocity of 32 ft/s. the height of this stone above the ground t s seconds after it is thrown is f(t)=-16t^2+32t+48. if a second stone is thrown from the found, then its height above the ground after t seconds os given by g(t)=-16t^2+vt,where v is the initial velocity of the second stone. determine the value of v so that both stones reach the same height.
回答 (2)
2013-10-03 7:03 am
✔ 最佳答案
height of first stone :
f(t) = -16t^2 + 32t + 48
f'(t) = -32t + 32
f''(t) = -32 < 0 ; that means when f'(t) = 0, the function is a max.
when f'(t) = 0
0 = -32t + 32
t = 1
f(1) = -16(1)^2 + 32(1) + 48 = 64
g(t)=-16t^2+vt
g'(t) = -32t + v
when g'(t) = 0 ; let the time t be tm
0 = -32tm + v
v = 32tm
max. height of the second stone = 64
64 = -16tm^2 + (32tm) tm
64 = 16tm^2
tm = 2
v = 32tm = 64
2013-10-03 2:51 am
f(t)=-16t²+32t+48
f'(t)=-32t+32
f'(t)=0 => t=1
f''(t)=-32 < 0
max f(t)=f(1)=64
g(t)=-16t²+vt
g'(t)=-32t + v
g'(t)=0 => t = v/32
g''(t)=-32 < 0
max g(t) = g(v/32) = -16(v/32)²+v(v/32) = v²/32-v²/64 = v²/64
For max f(t)=max g(t), v²/64=64
v² = 64²
v = 64 (+ve)
f'(t)=-32t+32
f'(t)=0 => t=1
f''(t)=-32 < 0
max f(t)=f(1)=64
g(t)=-16t²+vt
g'(t)=-32t + v
g'(t)=0 => t = v/32
g''(t)=-32 < 0
max g(t) = g(v/32) = -16(v/32)²+v(v/32) = v²/32-v²/64 = v²/64
For max f(t)=max g(t), v²/64=64
v² = 64²
v = 64 (+ve)
收錄日期: 2021-04-16 16:16:42
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20131002000051KK00137