If an object is launched straight up into the air from a starting height of h_{0} feet, then the height of the object after t seconds is approximately h=-16t^2+v_{0}t+h_{0} feet, where v_{0} is the initial velocity of the object. Find the starting height and initial velocity of an object that attains a maximum height of 412 feet five seconds after being launched.

Answer 1

Did you not look how your previous two posts of the same question came out?

Hard to guess what you mean by
h_{0} and v_{0}t

That is why your question does not get answered

Answer 2

Knowing the formula, it seems clear that h_{0} means h₀.

So, forgetting the subscripts, let's just say

y = -16t^2 + vt + h
y' = -32t + v

y'=0 when t=5
-32(5) + v = 0
v = 160

y = -16t^2 + 160t + h
412 = -16(25)+160(5)+h
h = 12