you throw a ball upward. its height h in feet after t seconds can be modeled by the function h=-16t^2+30t+6.after how many seconds will it hit the ground
To find when the ball will hit the ground, we need to find the time when the height will be 0.
Setting h = 0 in the equation h = -16t^2 + 30t + 6:
0 = -16t^2 + 30t + 6
Rearranging this equation gives:
16t^2 - 30t - 6 = 0
Dividing by 2 gives:
8t^2 - 15t - 3 = 0
This is a quadratic equation in the form of at^2 + bt + c = 0, where a = 8, b = -15, and c = -3.
To solve this quadratic equation, you can use the quadratic formula:
t = [-b ± √(b^2 - 4ac)] / 2a
Plugging in the values of a, b, and c:
t = [15 ± √((-15)^2 - 4(8)(-3))] / 2(8)
t = [15 ± √(225 + 96)] / 16
t = [15 ± √321] / 16
t ≈ [15 ± 17.92] / 16
Now, we have two possible solutions:
t ≈ [15 + 17.92] / 16 ≈ 2.12 seconds
t ≈ [15 - 17.92] / 16 ≈ -0.52 seconds
Since time can't be negative, the ball will hit the ground after approximately 2.12 seconds.