A launched rocket has an altitude in meters, given by the polynomial h+vt-4.9^2, h is the height in meters v is the velocity in meters per second and t is the number of seconds for which it takes the rocket to become airborne. If the rocket is launched from the top of a tower that is 150 meters and the initial speed is 40 meters per second, what will its height be after 2 seconds rounded to the nearest tenth?
h = 150 + Vo*t - 4.9T^2.
h = 150 + 40*2 - 4.9*2^2,
h = 150 + 60.4 = 210.4m above ground.
h = 210.4 - 150 = 60.4m above the tower.
To find the height of the rocket after 2 seconds, we can substitute the given values into the polynomial equation.
In the given polynomial equation:
h = height of the rocket
v = velocity of the rocket
t = time in seconds
Given:
h = 150 meters (height of the tower)
v = 40 meters per second (initial speed)
t = 2 seconds (time elapsed)
Substituting these values into the polynomial equation:
h + vt - 4.9t^2
= 150 + (40 * 2) - 4.9 * (2)^2
= 150 + 80 - 4.9 * 4
= 150 + 80 - 4.9 * 16
= 150 + 80 - 78.4
= 230 - 78.4
= 151.6
Therefore, the height of the rocket after 2 seconds, rounded to the nearest tenth, is approximately 151.6 meters.