If an object is dropped from 6 feet above the the ground, its height h(t) after t seconds can be modeled by a quadratic equation

Question

If an object is dropped from 6 feet above the the ground,
its height h(t) after t seconds can be modeled by a quadratic equation.

 

 

t (seconds) can be assigned the number 1 in this instance.

What is needed is a equation made from this model. 

Answer 1

h(t) = -16t^2 + 6

general formula is

h(t) = -16t^2 + vot + ho where t=time, vo = initial velocity, ho=initial height

vo apparently is zero in this case, ho = 6

the -16t^2 term is deceleration due to gravity, in feet per second per second

if t=1

then

h(1) = -16(1)^2 + 6 = 16+6 = -10. that would seem to mean there's a hole and the object goes 10 feet down the hole, below ground