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. In addition, the function d(t) = 85t models the horizontal distance d traveled by the ball. 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.

h(t)=0 when t = 1.5

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

The height of the ball is given by the equation h(t) = -16t^2 + 22t + 3.

Setting h(t) equal to zero gives us:

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

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

t = (-b ± √(b^2 - 4ac)) / (2a)

For this equation, a = -16, b = 22, and c = 3. Plugging these values into the quadratic formula, we can solve for t:

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

Simplifying further, we have:

t = (-22 ± √(484 + 192)) / (-32)

t = (-22 ± √676) / -32

t = (-22 ± 26) / -32

Now we can calculate the two possible values of t:

t₁ = (-22 + 26) / -32 = 4 / -32 = -0.125

t₂ = (-22 - 26) / -32 = -48 / -32 = 1.5

Since time cannot be negative, we discard the negative solution, leaving us with t = 1.5.

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

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

Given that the height h(t) of the ball is modeled by the equation h(t) = -16t^2 + 22t + 3, we can set h(t) equal to zero and solve for t.

So, we have -16t^2 + 22t + 3 = 0.

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

t = (-b ± √(b^2 - 4ac))/(2a)

Applying this formula to our equation, where a = -16, b = 22, and c = 3, we get:

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

Simplifying this further, we have:

t = (-22 ± √(484 + 192))/(2(-16))
t = (-22 ± √(676))/(2(-16))

Now, we take the square root of 676, which is 26:

t = (-22 ± 26)/(2(-16))

Simplifying this expression, we have:

t = (-22 ± 26)/(-32)

Now, we can calculate the values of t:

t1 = (-22 + 26)/(-32) = 4/-32 = -1/8 ≈ -0.125
t2 = (-22 - 26)/(-32) = -48/-32 = 3/2 = 1.5

Since time cannot be negative, we can disregard the first solution (t1 ≈ -0.125).

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