# The height h in feet of an object after t seconds is given by the function

h=-16t^2+70t+5. How long will it take the object to hit the ground? Round your answer to the nearest thousandth.

If that is the function, you need to have a value for the height to find the time.

I hope this helps a little. Thanks for asking.

## To determine when the object will hit the ground, we need to find the value of t when the height h is equal to 0.

The given function is:

h = -16t^2 + 70t + 5

Setting h = 0, we have:

0 = -16t^2 + 70t + 5

t = (-b ± √(b^2 - 4ac)) / (2a)

In our case, a = -16, b = 70, and c = 5. Plugging these values into the formula:

t = (-(70) ± √((70)^2 - 4(-16)(5))) / (2(-16))

Simplifying further:

t = (-70 ± √(4900 + 320)) / -32
t = (-70 ± √5220) / -32

Using a calculator or computer software, we can approximate the square root:

t ≈ (-70 ± 72.257) / -32

Now, we can solve for the positive and negative cases:

t1 = (-70 + 72.257) / -32
t1 ≈ 0.071 seconds

t2 = (-70 - 72.257) / -32
t2 ≈ 4.571 seconds

Since time cannot be negative, the object will hit the ground approximately 0.071 seconds after it is launched.

Therefore, the object will take approximately 0.071 seconds to hit the ground.

## To find the time it takes for the object to hit the ground, we need to find the time when the height is equal to zero. In other words, we need to solve the equation -16t^2 + 70t + 5 = 0.

To solve this quadratic equation, we can use the quadratic formula: t = (-b ± √(b^2 - 4ac)) / (2a), where a, b, and c are the coefficients of the quadratic equation.

In this case, the coefficients are:
a = -16
b = 70
c = 5

Substituting these values into the quadratic formula, we get:
t = (-70 ± √(70^2 - 4(-16)(5))) / (2(-16))

Simplifying further, we have:
t = (-70 ± √(4900 + 320)) / (-32)
t = (-70 ± √(5220)) / (-32)

The square root of 5220 is approximately 72.166, so:
t = (-70 ± 72.166) / (-32)

Now we have two possible values for t, taking both the positive and negative square root:
t1 = (-70 + 72.166) / (-32) ≈ 0.07
t2 = (-70 - 72.166) / (-32) ≈ 4.47

Since time cannot be negative in this context, we discard t2. Therefore, it will take approximately 0.07 seconds (or 0.070 to the nearest thousandth) for the object to hit the ground.