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}.
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?
1 answer
1 answer