To find the time when the softball will be 50 feet above the ground, we need to set the height equation equal to 50 and solve for t.
-16t^2 + 70t + 3 = 50
Subtracting 50 from both sides:
-16t^2 + 70t - 47 = 0
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = -16, b = 70, and c = -47.
t = (-70 ± √(70^2 - 4*(-16)*(-47))) / 2*(-16)
t = (-70 ± √(4900 - 3016)) / (-32)
t = (-70 ± √(1884)) / (-32)
Calculating the square root of 1884, we get:
t ≈ (-70 ± 43.39) / (-32)
Now we have two solutions:
1) t = (-70 + 43.39) / (-32) ≈ (-26.61) / (-32) ≈ 0.831 seconds
2) t = (-70 - 43.39) / (-32) ≈ (-113.39) / (-32) ≈ 3.543 seconds
Therefore, the time when the softball will be closest to 50 feet above the ground is approximately 0.831 seconds.
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 left-parenthesis t right-parenthesis equals negative 16 t squared plus 70 t plus 3, where t is time in seconds and t greater than or equal to 0. Which is closest to the time when the softball will be 50 feet above the ground?
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