A baseball player hits a baseball into the outfield. The following equation models the ball's path. If not caught, when will the baseball land? Round to the nearest tenth second.

h = -16t^2 + 50t + 4

OK. The four questions you posted are all basically the same. Here's help on this one:

The baseball will land when the height is zero.

So, you need to solve

-16t^2 + 50t + 4 = 0

Unless you see an abvikus factoring, use the quadratic formula:

for f(x) = ax^2 + bx + c,

x = (-b ± √(b^2 - 4ac)]/2

Here, a = -16
b = 50
c = 4

t = (-50 ± √(50^2 - 4(-16)(4))]/-32
= (-50 ± √(2500 + 256)]/-32
= (-50 ± √2756]/-32
= (-50 ± 52.5)]/-32

t = -0.07 or 3.2

Since time is positive, use t = 3.2 sec

To find when the baseball will land, we need to determine the value of 't' when the height 'h' is equal to zero since the ball reaches the ground when its height is zero.

The given equation is h = -16t^2 + 50t + 4.

Setting h to zero, we have 0 = -16t^2 + 50t + 4.

Now, we need to solve this quadratic equation to find the values of 't' when the height is zero.

To solve the quadratic equation, we can use the quadratic formula: t = (-b ± √(b^2 - 4ac)) / (2a), where 'a', 'b', and 'c' are the coefficients in the quadratic equation.

In this case, 'a' = -16, 'b' = 50, and 'c' = 4.

Substituting these values into the quadratic formula, we get:

t = (-50 ± √(50^2 - 4*(-16)*4)) / (2*(-16))

Simplifying this equation will give us two possible values for 't'.

t₁ = (-50 + √(2500 - (-256))) / (-32)
t₂ = (-50 - √(2500 - (-256))) / (-32)

Evaluating these equations, we get:

t₁ = (-50 + √(2500 + 256)) / (-32)
t₂ = (-50 - √(2500 + 256)) / (-32)

t₁ ≈ 0.9 seconds
t₂ ≈ 3.2 seconds

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

Therefore, the baseball will land approximately 0.9 seconds after it was hit into the outfield.