A launched rocket has an altitude, in meters,give n by the polynomial h+vt-4.9t^2, where h is the height, in meters, from which the rocket is airborne. If a rocket is launched from the top of the tower 90 meters high with an initial upward speed of 50 meters per second, what will its height be after 2 seconds.

height = -4.9t^2 + 50t + 90

sub in t=2, and evaluate

To find the height of the rocket after 2 seconds, we need to substitute the given values into the equation and solve for the height (h).

Given:
Initial height, h = 90 meters
Initial upward speed, v = 50 meters per second
Time, t = 2 seconds

The equation for the altitude of the rocket is h = vt - 4.9t^2.

Substituting the given values:
h = (50 * 2) - (4.9 * 2^2)

Simplifying:
h = 100 - (4.9 * 4)

h = 100 - 19.6

h = 80.4

Therefore, the height of the rocket after 2 seconds is 80.4 meters.