Question 1074061
the problem uses feet, so we will use 16 feet per second per second which is derived from the force of gravity
:
the formula for height at a give time is
:
h(t) = -16t^2 + v(0)t + h(0)
:
Note that h(0) is 0 at ground level
:
the graph of h(t) is a parabola that open downward
:
we are given v(0) = 80 feet per second
:
h(t) = -16t^2 +80t
:
we want to find the vertex of the parabola by taking the first derivative of h(t) and setting it = 0 and solve for t
:
h'(t) = -32t + 80
:
-32t = -80
:
t = 2.5 seconds
:
this tells us that the maximum height of the projectile occurs at 2.5 seconds after launch
:
h(t) = -16(2.5)^2 + 80(2.5) = 100 feet
:
The maximum height is 100 feet at 2.5 seconds after launch
:
now we look for the time after launch when the projectile reaches 96 feet
:
96 = -16t^2 + 80t
:
16t^2 -80t +96 = 0 
:
divide both sides of = by 16
:
t^2 -5t +6 = 0
:
factor the quadratic
:
(t-3) ^ (t-2) = 0
:
t is 3 or 2
:
Note the max height occurs at 2.5 seconds
:
**********************************************************************
The projectile's height exceeds 96 feet during the time interval (2, 3)
:
here is the graph of the projectiles path
:
{{{ graph( 300, 200, -1, 5, -1, 100, -16x^2 +80x) }}}
***********************************************************************
: