The height of an launched from the ground after t seconds is given by h(t)=-16t^2+32t. How long for the object to obtain a height of 32ft. Hit the ground?

part one, height = 32

-16t^2 + 32t = 32
t^2 - t +1 = 0

by the formula:
t = (1 ± √-3)/2 , not real
so it will never reach a height of 32 ft

we can see that by noting that the x-intercepts are 0 and 2
so the vertex is at t = 1
max height is h(1) = -16+32 = 16

I also incidentally answered the 2nd part of your question.

hit gourd?

To find the time it takes for the object to obtain a height of 32ft, we need to set the equation equal to 32 and solve for t.

Given:
h(t) = -16t^2 + 32t
We set h(t) equal to 32:

32 = -16t^2 + 32t

To find the time it takes for the object to hit the ground, we need to set h(t) equal to 0:

0 = -16t^2 + 32t

Now let's solve each equation separately.

First, solve for t when h(t) = 32:

32 = -16t^2 + 32t

Rearrange the equation:

16t^2 - 32t + 32 = 0

Divide through by 16 to simplify:

t^2 - 2t + 2 = 0

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

t = (-b ± √(b^2 - 4ac)) / 2a

In our case, a = 1, b = -2, and c = 2. Plugging these values into the formula:

t = (-(-2) ± √((-2)^2 - 4(1)(2))) / (2(1))

Simplifying further:

t = (2 ± √(4 - 8))/2

t = (2 ± √(-4))/2

Since the discriminant (b^2 - 4ac) is negative, the equation has no real solutions. Therefore, there is no real value of t for which the object reaches a height of 32ft.

Next, let's solve for t when h(t) = 0:

0 = -16t^2 + 32t

Rearrange the equation:

16t^2 - 32t = 0

Factor out 16t from both terms:

16t(t - 2) = 0

By the zero product property, we can set each factor equal to zero:

16t = 0 or t - 2 = 0

Solving each equation separately:

For 16t = 0:
t = 0

For t - 2 = 0:
t = 2

Therefore, the object hits the ground at t = 0 seconds (initial launch) and t = 2 seconds.

To find the time it takes for the object to reach a height of 32ft, we need to solve the equation h(t) = 32.

Given that h(t) = -16t^2 + 32t, we can substitute this into the equation and solve for t:

-16t^2 + 32t = 32

Rearranging the terms, we get:

-16t^2 + 32t - 32 = 0

Now, let's solve this quadratic equation. We can either factor or use the quadratic formula.

Using factoring:
-16t^2 + 32t - 32 = 0
-16(t^2 - 2t + 2) = 0
(t - 2)(t - 1) = 0

The solutions are t = 2 and t = 1.

This means that for the object to reach a height of 32ft, it takes 2 seconds and 1 second.

To find out when the object hits the ground, we need to set h(t) = 0 and solve for t.

-16t^2 + 32t = 0

Factoring out t, we get:

t(-16t + 32) = 0

t = 0 or -16t + 32 = 0

Simplifying the second equation, we have:

-16t + 32 = 0
-16t = -32
t = 2

So, the object hits the ground after 2 seconds.

Therefore, it takes 2 seconds for the object to reach a height of 32ft and to hit the ground.