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vicada2782
23.04.2020 •
Mathematics
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
Solved
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Ответ:
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
Ответ:
answer:
f
step-by-step explanation: