SOLUTION: I would really appreciate it if I could get some help on this question, thanks. The path of a falling object is given by the function s=-16t^2+vo+s0 where vo represents the init

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: I would really appreciate it if I could get some help on this question, thanks. The path of a falling object is given by the function s=-16t^2+vo+s0 where vo represents the init      Log On


   

Question 137457: I would really appreciate it if I could get some help on this question, thanks.
The path of a falling object is given by the function s=-16t^2+vo+s0 where vo represents the initial velocity in ft/sec and so represents the initial height in feet. Also, s represents the height in feet of the object at any time,t,which is measured in seconds.
If a rock is thrown upward with the initial velocity of 40ft per second from the top of a 30ft building.
s= -16t^2+vot+so I figured out that much, then it asks after how many seconds will the graph reach maximum height?What is the maximum height? I just need help setting up the equation to solve for height.... The question comes from my online algebra class at aiu. HELP PLEASE. thanks, alaska
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Given:
s%28t%29+=+-16t%5E2%2Bv%5B0%5Dt%2Bh%5B0%5D and...
v%5B0%5D+=+40
h%5B0%5D+=+30
Find the time, t, of the maximum height in seconds and the maximum height in feet.
Substitute the given values of v%5B0%5D+=+40 and h%5B0%5D+=+30 into the equation.
s%28t%29+=+-16t%5E2%2B40t%2B30
To find the value of t at the maximum height, you are really looking for the t-coordinate of the vertex of the parabola described by this quadratic equation.
This is given by:
t+=+-b%2F2a
The a and b come from the general form of the quadratic equation:ax%5E2%2Bbx%2Bc+=+0 but in your problem, the independent variable is t rather than x, where a = -16 and b = 40, so...
t+=+-40%2F2%28-16%29
t+=+40%2F32
t+=+5%2F4
t+=+1.25 seconds.
The maximum height is reached in 1.25 seconds.
To find the maximum height, just substitute t = 1.25 into the equation and solve for s.
s%281.25%29+=+-16%281.25%29%5E2%2B40%281.25%29%2B30
s%281.25%29+=+-16%281.5625%29%2B50%2B30
s%281.25%29+=+-25%2B80
s%281.25%29+=+55feet.
The maximum height reached is 55 feet.
graph%28400%2C400%2C-3%2C4%2C-5%2C60%2C-16x%5E2%2B40x%2B30%29