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?
Hmm -- you've posted this again with the same missing word.
Please proofread, correct, and repost.
Besides that , it was already answered.
To find the time it takes for the object to reach a height of 32ft, we need to solve the equation h(t) = 32.
The given equation for the height of the object is:
h(t) = -16t^2 + 32t
Substituting 32ft for h(t), we get:
-16t^2 + 32t = 32
Simplifying the equation:
-16t^2 + 32t - 32 = 0
Dividing the entire equation by -16 to simplify further:
t^2 - 2t + 2 = 0
Now, we can solve this quadratic equation for t. There are a couple of ways to do it, but let's use the quadratic formula:
The quadratic formula states:
For an equation of the form ax^2 + bx + c = 0, the solutions are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our equation, a = 1, b = -2, and c = 2, so we can plug these values into the formula to find t:
t = (-(-2) ± √((-2)^2 - 4(1)(2))) / (2(1))
t = (2 ± √(4 - 8)) / 2
t = (2 ± √(-4)) / 2
The discriminant (b^2 - 4ac) is negative in this case, which means there are no real solutions for t. This implies that the object will never reach a height of 32ft.
However, to determine how long it takes for the object to hit the ground, we need to find the time when h(t) = 0. This represents the point where the object touches the ground.
Equating h(t) to 0:
-16t^2 + 32t = 0
Factoring out common terms, we get:
-16t(t - 2) = 0
Using the zero-product property, we set each factor equal to zero:
-16t = 0 or t - 2 = 0
Solving each equation separately:
For -16t = 0, we have t = 0.
For t - 2 = 0, we have t = 2.
Therefore, it takes 2 seconds for the object to hit the ground.