a few algebra problems!? PLEASE help?
An object is launched upward at 45ft/sec from a platform that is 40 ft high. What is the objects maximum height if the equation of height (h) in terms of time (t) of the object is given by h(t) = -16t^2 + 45t + 40?
Comments
h(t) = -16t^2 + 45t + 40
taking derivative h'(t) = -32t + 45 and h'(t) = 0 gives t = 45/32 sec
as the time it takes to reach the maximum height. Substituting this in the height equation gives:
h(45/32) = -16*(45/32)² + 45*(45/32) + 40 = 71.640625 ft.
Try to use c- (b^2)/4a like the formula of h in the vertex (h,k)
h=c- (b^2)/4a
substitute the values of a,b,c from the equation
40 - (45^2)/[4(-16)]
which is equal to about 71.64
h'(t) = -32t + 45
Set that to zero:
-32t + 45 = 0
32t = 45
t = 45/32
t = 1.40625
h(1.40625) = -16(1.40625)^2 + 45(1.40625) + 40
h(1.40625) = -16(1.9775390625) + 63.28125 + 40
h(1.40625) = -31.640625 + 63.28125 + 40
h(1.40625) = 71.640625
That's about 71'8"
See also http://www.wolframalpha.com/input/?i=maximize+h%28...
dh/dt=32t+45
=32*45+45
=1485