An object is launched at 29.4 meters per

second (m/s) from a 34.3-meter tall

platform. The equation for the object's

height s at time x seconds after launch is

f (x) = - 4.9x2 + 29.4x + 34.3, where y is in

meters. What is the initial height of the

object?

29.4 meters

78.4 meters

34.3 meters

3 meters

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