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.