Question

Help I will give you brainliest if you do

Answer 1

Answer:

Step-by-step explanation:

If the position function is

$h(t)=-40t^2+200t$ and we are looking for time when the height is 0, we sub in a 0 for h(t) and solve for t:

$0=-40t^2+200t$ and the easiest way to do this is to factor by taking the GCF of -40t:

0 = -40t(t - 5) and by the Zero Product Property,

-40t = 0 or t - 5 = 0. Solving for t, we get

t = 0 (which is before the object is launched) and

t = 5 (which is how long it takes the object to go from the ground, up to its max height, and then back to the ground again).

Your choice is c) 5