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

for this case we have by definition, if two lines are perpendicular, then the product of their slopes is -1.

then, given the following line:

so,

we are looking for m_ {2}:

then, the line is given by:

we find "b" replacing the given point:

finally, the equation is:

option c

yes the data was not prsented

step-by-step explanation: