Asked by Maddie
You toss a rock up vertically at an initial speed of 55 feet per second and release it at an initial height of 6 feet. The rock will remain in the air for_______ seconds.
It will reach a maximum height of_______ feet after ________ seconds.
It will reach a maximum height of_______ feet after ________ seconds.
Answers
Answered by
Reiny
from the given data:
h = -16t^2 + 55t + 6
this is a parabola, we need its vertex.
the t of the vertex is -b/(2a) = -55/-32
= 55/32 seconds
plug that into equation to find max height
it is in the air for h > 0
let's see when h = 0
0 = -16t^2 + 55t + 6
16t^2 - 55t - 6 = 0
solve using the formula, since this does not factor.
ignore the negative root, just use the positive one
so the rock will be in the air for
0 > t > (above positive answer for t)
h = -16t^2 + 55t + 6
this is a parabola, we need its vertex.
the t of the vertex is -b/(2a) = -55/-32
= 55/32 seconds
plug that into equation to find max height
it is in the air for h > 0
let's see when h = 0
0 = -16t^2 + 55t + 6
16t^2 - 55t - 6 = 0
solve using the formula, since this does not factor.
ignore the negative root, just use the positive one
so the rock will be in the air for
0 > t > (above positive answer for t)
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