A player throws a ball up and toward a wall that is 17 feet high. The height h in feet of the ball after t seconds after it leaves the player's hand is modeled by h=-16t^2+25t+6. If the ball makes it to where the wall is, will it go over the wall or hit the wall?

The maximum height of the ball is reached at t=25/32

Plug that in. If it is greater than 17, then I guess it would depend on how far away the wall is.

To determine whether the ball will go over the wall or hit the wall, we need to find the time at which the ball reaches a height of 17 feet.

Given the equation h = -16t^2 + 25t + 6, we set h equal to 17 and solve for t:

17 = -16t^2 + 25t + 6

Rearranging the equation, we get:

16t^2 - 25t - 11 = 0

To solve this quadratic equation, we can use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / 2a

For our equation, a = 16, b = -25, and c = -11.

Substituting these values into the quadratic formula:

t = (-(-25) ± √((-25)^2 - 4 * 16 * -11)) / (2 * 16)

Simplifying further:

t = (25 ± √(625 + 704)) / 32

t = (25 ± √1329) / 32

Now we have two possible values for t. Let's calculate them:

t₁ = (25 + √1329) / 32 ≈ 1.825

t₂ = (25 - √1329) / 32 ≈ -0.381

Since time cannot be negative in this context, we discard t₂. Therefore, the only valid value for t is t₁.

The positive value of t represents the time it takes for the ball to reach a height of 17 feet. Based on this calculation, we can conclude that the ball will hit the wall when it reaches a height of 17 feet.