a ball is thrown downward from the too of 110 ft bldg with an initial velocity of 14ft per second. the height of ball h after t seconds is given by the equation h = -16t^2-14t +110.

how long after ball is thrown will it hit the ground

Look at the response to your previous post of this same question.

To find the time at which the ball hits the ground, we need to determine when the height of the ball (h) becomes zero.

We can set the equation for height (h) equal to zero and solve for t:
-16t^2 - 14t + 110 = 0

To solve this quadratic equation, we can either factor it (if possible) or use the quadratic formula. In this case, factoring might be a bit complicated, so we'll use the quadratic formula:

The quadratic formula is: t = (-b ± √(b^2 - 4ac)) / 2a

Let's assign values to the respective variables in the quadratic formula:
a = -16
b = -14
c = 110

Substituting these values into the formula:

t = (-(-14) ± √((-14)^2 - 4(-16)(110))) / (2(-16))
t = (14 ± √(196 + 7040)) / (-32)
t = (14 ± √7236) / (-32)

Now, let's calculate the values within the square root:
√7236 ≈ 85.06

So we have two possible values for t:
t₁ = (14 + 85.06) / (-32)
t₂ = (14 - 85.06) / (-32)

Evaluating these expressions:
t₁ ≈ 2.85
t₂ ≈ -3.21

Since time can only be positive, the ball will hit the ground approximately 2.85 seconds after it is thrown.