Question

When an object is thrown upward with an

initial velocity of 96 ft/sec from a height of
256 ft, its height h after t seconds is given
by the function below.
h(t)= - 16t² +96t+256
UR
D
256 ft-
h(t)=16t+961 +256
h(t)
What is the height of the object after 0 sec. 1 sec, 3 sec, 4 sec, and 6 sec?
The height of the object is ft.
(Type each value. in order. Use a comma to separate answers as needed.)

Answer 1

Explanation:

knowing the physics principles and how the formulas work, the function must be

h(t) = -16t² + 96t + 256

what you did then with the second line of h(t) definition - only God knows.

anyway, to get the height of the object after x seconds, we only need to put the value of x into every place of t in the function expression and simply calculate. that is how functions work. any and every type of function.

0 seconds. t = 0

h(0) = -16×0² + 96×0 + 256 = 256ft

1 second. t = 1

h(1) = -16×1² + 96×1 + 256 = -16 + 96 + 256 = 336ft

3 seconds. t = 3

h(3) = -16×3² + 96×3 + 256 = -144 + 288 + 256 = 400ft

4 seconds. t = 4

h(4) = -16×4² + 96×4 + 256 = -256 + 384 + 256 = 384ft

6 seconds. t = 6

h(6) = -16×6² + 96×6 + 256 = -576 + 576 + 256 = 256ft

0 seconds 256 ft

1 second 336 ft

3 seconds 400 ft

4 seconds 384 ft

6 seconds 256 ft

so, after 6 seconds it is back to the same height as what it started from.