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
To solve for t when the object hits the ground (h=0), we can substitute h=0 into the equation and solve it as a quadratic equation. So we need to fill in the blanks of the quadratic formula.
The quadratic formula is given by:
t = (-b ± √(b^2 - 4ac)) / (2a),
where a, b, and c are the coefficients of the quadratic equation h=-16t^2+vt+s. In this case, a=-16, b=v, and c=s.
Filling in the blanks of the quadratic formula, we have:
Blank 1: Since there are only two possibilities of sign in the quadratic formula, we can choose either "+" or "-" because both will result in a valid solution.
Blank 2: Calculate the value inside the square root (√):
Blank 2 = b^2 - 4ac
Blank 3: After calculating Blank 2, fill in the value you obtained in Blank 2 here.
So, the quadratic formula becomes:
t = (-v ± √(v^2 - 4(-16)(s))) / (2(-16))
Please note that the earlier expression for h=-16t^2+vt+s assumes the object is thrown upwards with an initial velocity of v.
To find the time when the object hits the ground, we set h = 0 in the equation:
0 = -16t^2 + vt + s
This is a quadratic equation of the form ax^2 + bx + c = 0, where:
a = -16
b = v
c = s
Now, we can substitute these values into the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Replacing the blanks with the corresponding values:
blank 1 = -v
blank 2 = v^2 - 4(-16)(s)
blank 3 = -v
Therefore, the quadratic formula becomes:
t = (-v ± √(v^2 - 4(-16)(s))) / (2(-16))