Question 1150654
An object is projected directly upward from an initial height of 24 feet at an initial velocity of 96 feet per second.
 Write a model for the height of the object, s(t), after t seconds.
 Use s(t) = -16t^2+v+s.
v=96, s=24, therefore
s(t) = -16t^2 + 96t + 4
:
After how many seconds does the object reach its maximum height?
Max occurs at the axis of symmetry, (x = -b/2a), 
in this equation a=-16, b=96 therefore
t = {{{(-96)/(2*-16)}}}
t = +3 sec for max height
:
 What is the maximum height?
s(t) = the height, t=3 therefore
s)t) = -16(3^2) + 96(3) + 24
s(t) = -144 + 288 + 24
s(t) = 168 ft is the max height
:
 After how many seconds does the object land on the ground?
then the height = 0, therfore
-16t^2 + 96t + 24 = 0
simplify divide by -4
4t - 24t - 6 = 0
Use the quadratic formula, a=4, b=-24, c=-6
the positive solution is
t ~ 6.24 seconds to reach the ground