During a softball game, a batter hits a ball upward from an initial height of 3 feet. The height, in feet, of the softball is given by
s(t)=-16t^2+70t+3 where t is time in seconds and t>=0. Which is closest to the time when the softball will be 50 feet above the ground?
To find the time when the softball will be 50 feet above the ground, we need to set s(t) equal to 50 and solve for t.
So, we have:
-16t^2 + 70t + 3 = 50
-16t^2 + 70t - 47 = 0
To solve this quadratic equation, we can use the quadratic formula, which states that for an equation of the form ax^2 + bx + c = 0, the solutions are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = -16, b = 70, and c = -47. Plugging these values into the quadratic formula, we get:
t = (-70 ± √(70^2 - 4(-16)(-47))) / (2(-16))
t = (-70 ± √(4900 - 3016)) / (-32)
t = (-70 ± √(1884)) / (-32)
t ≈ (-70 ± 43.4) / (-32)
The two possible values of t are:
t ≈ (-70 + 43.4) / (-32) ≈ 1.27 seconds
t ≈ (-70 - 43.4) / (-32) ≈ 3.95 seconds
The closest time when the softball will be 50 feet above the ground is approximately 1.27 seconds. Answer: \boxed{1.27}.