A baseball player hits a ball up in the air with an upward velocity of 18 ft/sec. The equation that gives the height of the ball at any time "t" is h(t) = -8t(t) + 18t +5. How long did it take for the ball to hit the ground?
Sorry I don't know how to type t squared.
No problem! I understand that you meant to type "t^2" when referring to "t" squared.
To find out how long it took for the ball to hit the ground, we need to determine the time when the height, h(t), is equal to zero. This is because at that point, the ball will have reached the ground.
Given the equation for the height of the ball at any time, h(t) = -8t^2 + 18t + 5, we can set h(t) equal to zero and solve for "t":
-8t^2 + 18t + 5 = 0
To solve this quadratic equation, you can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = -8, b = 18, and c = 5. Plugging these values into the quadratic formula, we get:
t = (-18 ± √(18^2 - 4(-8)(5))) / (2(-8))
Simplifying further:
t = (-18 ± √(324 + 160)) / (-16)
t = (-18 ± √484) / (-16)
t = (-18 ± 22) / (-16)
This gives us two possible solutions:
t = (18 + 22) / 16 = 40 / 16 = 2.5
t = (18 - 22) / 16 = -4 / 16 = -0.25
Since time cannot be negative in this context, we can disregard the negative value. Therefore, the ball takes 2.5 seconds to hit the ground.