Asked by Keonn'a
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?
Answers
Answered by
Reiny
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.
-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.
Answered by
Anonymous
hit gourd?
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