During softball practice, a softball pitcher throws a ball whose height can be modeled by the equation h = -16t^2 (<--squared) + 24t + 1, where h=height in feet and t = time in seconds. How long does it take for the ball to reach a height of 6 feet?

just plug in the numbers:

6 = -16t^2 + 24t + 1
16t^2 - 24t + 5 = 0
(4t-1)(4t-5) = 0

t = 1/4 (ball rising)
t = 5/4 (ball falling)

0.25 secs,1.25 secs

Given, height = 6 = -16
t
2
+ 24t + 1


16
t
2
+ 24t + 5 = 0


(4t-1) (4t-5) = 0

t =
1
4
,
5
4
or 0.25,1.25

It takes 0.25 secs and 1.25secs for the ball to go up and down respectively.

To find the time it takes for the ball to reach a height of 6 feet, we can set the equation equal to 6 and solve for t.

h = -16t^2 + 24t + 1

Setting h = 6:

6 = -16t^2 + 24t + 1

Combining like terms:

-16t^2 + 24t - 5 = 0

This is a quadratic equation in standard form, with a = -16, b = 24, and c = -5. We can use the quadratic formula to solve for t.

The quadratic formula is:

t = (-b ± √(b^2 - 4ac)) / 2a

Substituting the values:

t = (-24 ± √(24^2 - 4(-16)(-5))) / (2(-16))

Simplifying:

t = (-24 ± √(576 - 320)) / (-32)

t = (-24 ± √256) / (-32)

t = (-24 ± 16) / (-32)

Now we have two possible solutions:

t1 = (-24 + 16) / (-32) = -8 / -32 = 0.25

t2 = (-24 - 16) / (-32) = -40 / -32 = 1.25

Therefore, it takes approximately 0.25 seconds and 1.25 seconds for the ball to reach a height of 6 feet.

To find the time it takes for the ball to reach a height of 6 feet, we need to set the equation h = 6 and solve for t.

The equation given is: h = -16t^2 + 24t + 1

Substituting h with 6, the equation becomes: 6 = -16t^2 + 24t + 1

Now, let's rearrange this equation to solve for t.

Combine like terms: -16t^2 + 24t + 1 - 6 = 0

Simplify: -16t^2 + 24t - 5 = 0

To solve this quadratic equation, we can either factor it, complete the square, or use the quadratic formula. In this case, let's use the quadratic formula: t = (-b ± √(b^2 - 4ac)) / (2a)

For our equation, a = -16, b = 24, and c = -5. Plugging these values into the formula:

t = (-(24) ± √((24)^2 - 4(-16)(-5))) / (2(-16))

Now, let's simplify the equation:

t = (-24 ± √(576 - 320)) / (-32)

t = (-24 ± √256) / (-32)

Now, let's evaluate the expression under the square root:

t = (-24 ± 16) / (-32)

This gives two possible values for t:

t1 = (-24 + 16) / (-32) = -8 / (-32) = 1/4

t2 = (-24 - 16) / (-32) = -40 / (-32) = 5/4

Therefore, the ball reaches a height of 6 feet at two different times: 1/4 seconds and 5/4 seconds.