SOLUTION: The path of a falling object is given by using the function s= -16t^2+ v sub0 t+ s sub0 where v sub0 represents the initial height in ft/sec and s sub0 represents the initial h

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: The path of a falling object is given by using the function s= -16t^2+ v sub0 t+ s sub0 where v sub0 represents the initial height in ft/sec and s sub0 represents the initial h      Log On


   

Question 150685: The path of a falling object is given by using the function s= -16t^2+ v sub0 t+ s sub0 where
v sub0 represents the initial height in ft/sec and
s sub0 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.
A. If a rock is thrown upward with an initial velocity of 40 feet per second from the top of a 30 foot building, write the height (s) equation using this information...
I got.. -16t^2 + 40t + 30
B.How high is the rock after 2 seconds?
I got..
s(t)= -16(2)^2 + 40(2) + 30
s(t)= 46
C. After how many seconds will the graph reach maximum height? Show work algebraically.

D What is the maximum height?


I need help on C and D... and please tell me if I have figured A and B correctly... Thank you so very much.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Parts A and B are correct.

C)


From s%28t%29=-16t%5E2%2B40t%2B30, we can see that a=-16, b=40, and c=30.

In order to find out when the rock will reach the max height, use this formula: t=%28-b%29%2F%282a%29.


t=%28-b%29%2F%282a%29 Start with the given formula.


t=%28-%2840%29%29%2F%282%28-16%29%29 Plug in a=-16 and b=40.


t=%28-40%29%2F%28-32%29 Multiply 2 and -16 to get -32.


t=5%2F4 Reduce.

So the rock will reach the max height when t=5%2F4 or t=1.25. So the rock takes 1.25 seconds to reach the peak.

--------------------------
D)

s%28t%29=-16t%5E2%2B40t%2B30 Start with the given equation.


s%285%2F4%29=-16%285%2F4%29%5E2%2B40%285%2F4%29%2B30 Plug in t=5%2F4.


s%285%2F4%29=-16%2825%2F16%29%2B40%285%2F4%29%2B30 Square 5%2F4 to get 25%2F16.


s%285%2F4%29=-25%2B40%285%2F4%29%2B30 Multiply -16 and 25%2F16 to get -25.


s%285%2F4%29=-25%2B50%2B30 Multiply 40 and 5%2F4 to get 50.


s%285%2F4%29=55 Combine like terms.

So at the peak, the rock is 55 feet high. So the max height is 55 feet.