mathalgebra
posted by Michelle on .
A launched rocket has an altitude, in meters, given by the polynomial h = vt 4.9t^2, where h is the height, in meters, from which the launch occurs, at velocity v in meters per second, and t is the number of seconds for which the rocket is airborne. If a rocket is launched from the top of the tower 110 meters high with an initial upward speed of 60 meters per second, what its height be after 2 seconds?

If h is the height above ground level, then the equation will be
h = 4.9t^2 + 60t + 110
replace t with 2 and you got it. 
I got 422.08 but it says that is still wrong can you help

h = 4.9(4) + 60(2) + 110
= 210.4 m
or 100.4 above the tower