Question 189928
A baseball player hits a baseball into the air from a height of 3 feet with
 an initial vertical velocity of 72 feet per second. After how many seconds
 is the baseball 84 feet above the ground?
:
The equation for this is:
h = -16t^2 + 72t + 3
Where:
h = height at t seconds
-16t^2 is the downward pull of gravity
72t is the vertical velocity upward
3 is the initial height at t=0
:
Find t when height is 84 ft
-16t^2 + 72t + 3 = 84
Arrange as a quadratic equation:
-16t^2 + 72t + 3 - 84 = 0
-16t^2 + 72t - 81 = 0
multiply equation by -1 to change the signs, easier to factor
16t^2 - 72t + 81 = 0
this is a perfect square
(4t - 9)(4t - 9) = 0
4t = +9
t = {{{9/4}}}
t = 2.25 sec it will be at 84' (this is also the highest point (vertex))