Step-by-step explanation:
h(t) = -5t² + 20t + 18
this is a parabola (the curve an object follows when thrown up into the air and then falling back down).
and as such it will most likely have 2 solutions (one for when going up, and one for when falling back down again).
33 = -5t² + 20t + 18
15 = -5t² + 20t
-3 = t² - 4t
t² - 4t + 3 = 0
the solution of such a quadratic equation is
t = (-b ±sqrt(b² - 4ac))/(2a)
in our case
a = 1
b = -4
c = 3
t = (4 ±sqrt(16 - 4×1×3))/(2×1) = (4 ±sqrt(16-12))/2 =
= (4 ±sqrt(4))/2 = (4 ± 2)/2
t1 = (4+2)/2 = 6/2 = 3
t2 = (4-2)/2 = 2/2 = 1
the object reaches the height of 33 m first after 1 second (while going up), and then again after 3 seconds (while falling down again).
