If an object is thrown vertically upward with an initial velocity of v, from an original position of s, the height h at any time t is given by: h=-16t^2+vt+s (where h and s are in ft, t is in seconds and v is in ft/sec) Solve for t when the object hits the ground. (when h=0) t=blank 1 ±sqrt blank 2-4(blank3)/2(-16) fill in the blanks of the quadratic formula. Enter blank 1, blank 2, blank 3

Question

If an object is thrown vertically upward with an initial velocity of v, from an original position of s, the height h at any time t is given by: h=-16t^2+vt+s (where h and s are in ft, t is in seconds and v is in ft/sec) Solve for t when the object hits the ground. (when h=0) t=blank 1 ±sqrt blank 2-4(blank3)/2(-16) fill in the blanks of the quadratic formula. Enter blank 1, blank 2, blank 3

Answer 1

To solve for t when the object hits the ground, we need to find the roots of the quadratic equation h = -16t^2 + vt + s, where h = 0.

So, we can rewrite the equation as:
-16t^2 + vt + s = 0

Now, let's fill in the blanks of the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

In this case:
a = -16
b = v
c = s

Thus, the blanks can be filled as follows:
blank 1: -v
blank 2: v^2 - 4(-16)(s)
blank 3: -16

So, the quadratic formula becomes:
t = (-v ± √(v^2 - 4(-16)(s))) / (2(-16))

Now, you can substitute the values of v and s into the formula to get the answer.