Question

PLEASE HELP ME! I'LL GIVE BRAINLIEST TO WHOEVER GOT IT CORRECT!
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 heights at time t seconds after launch is $s(t) = -4.9t + 39.2t + 42.3$, where s is in meters. Create a table of values and graph the function. Approximately when will the object hit the ground?

Answer 1

Answer:

34.3 meters

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.