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 the ball hits the ground height be zero.
25 t - 16 t ^ 2 = 0
t ( 25 - 16 t ) = 0
This equation will be zero when t = 0
OR
25 - 16 t = 0
25 - 16 t = 0 Add 16 to both sides
25 - 16 t + 16 t = 0 + 16 t
25 = 16 t Divide both sides by 16
25 / 16 = t
t = 25 / 16
When t = 0 ball does not fly.
So solution are :
t = 25 / 16 seconds
To determine how many seconds go by before the ball hits the ground, we need to find the value of t when the height of the ball is zero. We can do this by solving the quadratic equation given by the equation height = 25t - 16t^2.
The quadratic equation is in the form ax^2 + bx + c = 0, where:
a = -16 (coefficient of t^2)
b = 25 (coefficient of t)
c = 0 (constant term)
We can use the quadratic formula to find the solutions for t:
t = (-b ± √(b^2 - 4ac)) / 2a
Plugging in the values for a, b, and c, we get:
t = (-(25) ± √((25)^2 - 4(-16)(0))) / 2(-16)
Simplifying further:
t = (-25 ± √(625)) / (-32)
Since we are interested in the time it takes for the ball to hit the ground, we only consider the positive solution:
t = (-25 + √625) / (-32)
t = (-25 + 25) / (-32)
t = 0 / (-32)
t = 0
Therefore, the ball hits the ground at t = 0 seconds.
Note: In this equation, it is assumed that the initial height of the ball is 0, and time is measured from the moment the ball is thrown in the air.