Respuesta :
For this case we have an equation of the form:
[tex]h (t) = (1/2) * (a) * (t ^ 2) + v0 * t + h0 [/tex]
Where,
h0: initial height
v0: initial speed
a: acceleration
Substituting values we have:
[tex]h (t) = (1/2) * (- 32) * (t ^ 2) + 0.25 * t + 12 [/tex]
Rewriting we have:
[tex]h (t) = -16t ^ 2 + 0.25 * t + 12 [/tex]
Note: see attached image
Answer:
[tex]h (t) = -16t ^ 2 + 0.25 * t + 12[/tex]
[tex]h (t) = (1/2) * (a) * (t ^ 2) + v0 * t + h0 [/tex]
Where,
h0: initial height
v0: initial speed
a: acceleration
Substituting values we have:
[tex]h (t) = (1/2) * (- 32) * (t ^ 2) + 0.25 * t + 12 [/tex]
Rewriting we have:
[tex]h (t) = -16t ^ 2 + 0.25 * t + 12 [/tex]
Note: see attached image
Answer:
[tex]h (t) = -16t ^ 2 + 0.25 * t + 12[/tex]

Answer:
it's D (the last graph)
Step-by-step explanation:
Because I got it right on EDG. and I did the math.