An object is launched at 39.2 meters per second (m/s) from a 42.3-meter tall platform. The equation for the object’s height s at time t seco

Question

An object is launched at 39.2 meters per second (m/s) from a 42.3-meter tall platform. The equation for the object’s height s at time t seconds after launch is where s is in meters. Create a table of values and graph the function. Approximately when will the object hit the ground?

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Maris 5 months 2021-08-15T07:07:18+00:00 1 Answers 1 views 0

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    2021-08-15T07:09:05+00:00

    Answer: Hello Luv………

    34.3 meters. Sorry I’m late.

    Step-by-step explanation:

    The generic equation for a movement with constant acceleration is:

    S = So + Vo*t + (a*t^2)/2

    Where S is the final position, So is the inicial position, Vo is the inicial speed, a is the acceleration and t is the time.

    If we compare with our equation (where x is the time and f(x) is the final distance), we have that:

    So = 34.3

    Vo = 29.4

    a = -9.8

    So we have that the inicial position (So) of the object is 34.3 meters.

    Hope this helps.

    Mark me brainest please.

    Anna ♥

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