A batter hits a baseball from 3eet above home plate along the path x=69t, y=3+40t-16t(squared.)
Incomplete.
81.125
To find the time at which the baseball hits the ground, we need to determine when the value of y becomes zero.
Given the equation y = 3 + 40t - 16t^2, we can set y to zero and solve for t:
0 = 3 + 40t - 16t^2
Rearranging the equation:
16t^2 - 40t - 3 = 0
We can solve this quadratic equation to find the values of t.
Using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 16, b = -40, and c = -3:
t = (-(-40) ± √((-40)^2 - 4 * 16 * -3)) / (2 * 16)
Simplifying:
t = (40 ± √(1600 + 192)) / 32
t = (40 ± √1792) / 32
t = (40 ± 42.35) / 32
For the two possible values of t, we have:
t1 = (40 + 42.35) / 32 ≈ 2.66 seconds
t2 = (40 - 42.35) / 32 ≈ -0.08 seconds
Since time cannot be negative in this context, we can discard t2. Therefore, the baseball will hit the ground approximately 2.66 seconds after being hit.