If an object is dropped from a height of 200 feet, the function h(t)=-16t^2+200 gives the height of the object after t seconds. Approximately, when will the object hit the ground? (1 point)
C. 3.54 seconds
To find the time when the object hits the ground, we need to find the value of t when the height, h(t), is equal to zero.
Given the function h(t) = -16t^2 + 200, we can set h(t) equal to zero and solve for t:
-16t^2 + 200 = 0
To solve this quadratic equation, we can factor it or use the quadratic formula. Factoring is not possible in this case, so we'll use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In the equation h(t) = -16t^2 + 200, we have a = -16, b = 0, and c = 200. Substituting these values into the formula:
t = (0 ± √(0^2 - 4(-16)(200))) / (2(-16))
Simplifying further:
t = ± √(0 + 12800) / -32
t = ± √12800 / -32
t = ± 112.8 / -32
We can ignore the negative solution (-112.8) since time cannot be negative in this context. Therefore, the object will hit the ground approximately 3.525 seconds after it is dropped.
The quadratic formula would also work here, right?
just solve for h=0.
-16t^2 + 200 = 0
16t^2 = 200
t^2 = 25/2
. . .