A baseball player hits a ball toward the outfield. The height h of the ball in feet is modeled by h(t) = -16t2 + 22t + 3, where t is the time in seconds. If no one catches the ball, how long will it stay in the air? (Round to the nearest tenth of a second and enter only the number.)

just solve for t when h=0.

It factors nicely, or your good old quadratic formula should do the trick.

Did you get 1.5 seconds?

To find how long the ball will stay in the air, we need to determine the time when the height of the ball is zero.

In the given model, the height of the ball is represented by the equation:
h(t) = -16t^2 + 22t + 3

To find when the ball is at ground level (height equals zero), we set h(t) to zero and solve for t:
0 = -16t^2 + 22t + 3

This is a quadratic equation in the form of at^2 + bt + c = 0, where
a = -16
b = 22
c = 3

We can solve this equation by factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)

Substituting the values of a, b, and c into the formula, we get:
t = (-22 ± √(22^2 - 4(-16)(3))) / (2(-16))

Simplifying further:
t = (-22 ± √(484 + 192)) / (-32)
t = (-22 ± √676) / (-32)
t = (-22 ± 26) / (-32)

By solving for both possibilities, we get:
t₁ = (-22 + 26) / (-32) = 4 / (-32) = -1/8 ≈ -0.1
t₂ = (-22 - 26) / (-32) = -48 / (-32) = 3/2 = 1.5

Since time cannot be negative in this context, we discard the negative value.

Therefore, the ball will stay in the air for approximately 1.5 seconds.

To find out how long the ball stays in the air, we need to determine when the height of the ball, represented by the function h(t), is equal to zero. This is because at the time when the height is zero, the ball will hit the ground.

The equation h(t) = -16t^2 + 22t + 3 models the height of the ball, with t representing time. We can set h(t) equal to zero and solve for t.

-16t^2 + 22t + 3 = 0

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

t = (-b ± √(b^2 - 4ac)) / (2a)

In our equation, a = -16, b = 22, and c = 3. Plugging these values into the quadratic formula, we have:

t = (-22 ± √(22^2 - 4(-16)(3))) / (2(-16))

Simplifying this expression further, we get:

t = (-22 ± √(484 + 192)) / (-32)

t = (-22 ± √676) / (-32)

t = (-22 ± 26) / (-32)

Now we have two possible values for t. Let's solve for both:

1. When t = (-22 + 26) / (-32), we get t = 1/8 or 0.125.
2. When t = (-22 - 26) / (-32), we get t = 3.

Since we are looking for the time when the ball hits the ground, we can discard the negative value of t, 0.125, because it represents an earlier point when the ball was still rising. Therefore, the ball will stay in the air for approximately 3 seconds.

Hence, the rounded answer is 3 seconds.