davisvictoria85
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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?

Sagot :

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 ♥

Answer:

The initial positio34.3 metres .

Step-by-step explanation:

The known equation for a movement

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

Where S is the final position, So is the initial position, Vo is the initial  speed, a is the acceleration due to gravity and t is the time taken.

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 m/s^2 because g act downside

Therefore the initial position (So) of the object is 34.3 meters.

Click here to know more about the equation for a movement

https://brainly.com/question/23571852

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