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
To find the height of the rocket after 2 seconds, we can substitute the given values into the equation for altitude.
The equation for altitude is h = vt - 4.9t^2, where h is the height, v is the velocity, and t is the time.
Given:
v = 60 meters per second
t = 2 seconds
Substituting these values into the equation, we get:
h = (60)(2) - 4.9(2)^2
First, let's simplify (2)^2:
h = (60)(2) - 4.9(4)
Now, calculate the values in parentheses:
h = 120 - 19.6
Finally, subtract 19.6 from 120 to find the height:
h = 100.4 meters
Therefore, the height of the rocket after 2 seconds is 100.4 meters.