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?

Question

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?

Answer 1

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.

Answer 2

I got 422.08 but it says that is still wrong can you help

Answer 3

h = -4.9(4) + 60(2) + 110

= 210.4 m

or 100.4 above the tower

Answer 4

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.