Asked by molly
A ball is thrown straight up. Its height, h (in metres), after t seconds is given by h = -5t2+10t+2. To the nearest tenth of a second, when is the ball 6m above the ground? Explain why there are two answers.
in depth answer please!
in depth answer please!
Answers
Answered by
mathhelper
Enter powers this way : -5^2
You want h to be 6, so
-5t^2+10t+2 = 6
5t^2 - 10t + 4 = 0
use formula since it doesn't factor
t = (10 ± √(100 -4(-5)(4)) )/10
= (10 ± √180)/10
= appr 2.34 or -.34
the corresponding parabola h = -5t^2 + 10t -4 has two x-intercepts, namely -.34 and 2.34
Since we only want h ≥ 0, we reject the negative answer.
The parabola doesn't know that it is to represent the height of an object, it
is merely just happy being a parabola.
You want h to be 6, so
-5t^2+10t+2 = 6
5t^2 - 10t + 4 = 0
use formula since it doesn't factor
t = (10 ± √(100 -4(-5)(4)) )/10
= (10 ± √180)/10
= appr 2.34 or -.34
the corresponding parabola h = -5t^2 + 10t -4 has two x-intercepts, namely -.34 and 2.34
Since we only want h ≥ 0, we reject the negative answer.
The parabola doesn't know that it is to represent the height of an object, it
is merely just happy being a parabola.
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