When a ball is thrown in the air, the height h of a ball in feet t seconds after the ball is thrown in the air, is given by the equation height = 25t–16t²
Use the quadratic equation to determine how many seconds goes by before the ball hits the ground?
when it hits the ground, height = 0
16t^2 -25t=0
t(16-25) = 0
t = 0 , (at the beginning of the throw,
or
t = 25/16 seconds
btw, if the quadratic equation ax^2 + bx + c = 0 has c = 0
then there is no need to use the formula, because it will always factor into x(ax + b) = 0
To determine how many seconds go by before the ball hits the ground, we need to find the time when the height, denoted by "h," is equal to zero.
In the given equation, height = 25t - 16t^2, we can set h = 0 and solve for t by using the quadratic equation.
The quadratic equation is given by t = (-b ± √(b² - 4ac)) / (2a), where the equation must be in the form at^2 + bt + c = 0.
In this case, we have a = -16, b = 25, and c = 0. Plugging these values into the quadratic equation, we have:
t = (-(25) ± √((25)² - 4(-16)(0))) / (2(-16)).
Simplifying further:
t = (-25 ± √(625)) / (-32).
t = (-25 ± 25) / (-32).
Therefore, we have two solutions:
1. t = (-25 + 25) / (-32) = 0 / (-32) = 0.
2. t = (-25 - 25) / (-32) = -50 / (-32) ≈ 1.5625.
Since time cannot be negative in this context, we can disregard the negative solution.
Therefore, it takes approximately 1.5625 seconds for the ball to hit the ground.