Aball is thrown from an initial height of 3 meters with an upward initial velocity of 25 m/s. the balls height h(in meters)after t seconds is given by the following. h=3+25t-5t^2, find all the values of t for which the balls height is 8 meters.

t ∈ {0.209, 4.791}

Step-by-step explanation:

Substituting 8 for h, we want to find t using ...

... 8 = 3 +25t -5t^2

... 5t^2 -25t +5 = 0 . . . . . subtract the right side to get standard form

... t^2 -5t +1 = 0 . . . . . . . . divide by 5 (makes the leading coefficient 1)

... (t -2.5)^2 -5.25 = 0 . . . rewrite to vertex form*

... t = 2.5 ± √5.25 . . . . . . add 5.25, square root, add 2.5

Rounded to milliseconds, the times are 209 ms and 4.791 s.

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* The vertex of this function is different from that of the graph because we have shifted it down 8, and removed a vertical scale factor of -5 relative to the one that is graphed.

is only a line? i think so

45

step-by-step explanation: