The height of a rocket in meters t seconds after it is launched is approximated by the polynomial 0.5 at2 + vt + y where a is always -9.8, v is the initial velocity, and h is the initial height. A 300X was launched from a height of 10 meters. What was its height after 3 seconds?
Height(t) = -4.9t^2 + vt + 10
so Height(3) = -4.9(9) + 3v+10
= 3v - 34.1
You will have to know v, the intital velocity, to find to find an actual numerical answer.
To find the height of the rocket after 3 seconds, we can substitute the given values into the polynomial equation and solve for the height.
Given:
a = -9.8 (acceleration due to gravity)
v = 300X (initial velocity)
h = 10 (initial height)
t = 3 (time after launch)
Substituting the values into the equation: h = 0.5at^2 + vt + h
h = 0.5(-9.8)(3)^2 + (300X)(3) + 10
Now, we can simplify the equation and calculate the height:
h = 0.5 * (-9.8) * 9 + 900X + 10
h = -44.1 + 900X + 10
h = 900X - 34.1
Thus, the height of the rocket after 3 seconds is given by the equation h = 900X - 34.1.