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?
Check your 5-15-11,2:27pm post for solution.
To find the height of the rocket after 2 seconds, we need to substitute the given values into the polynomial equation h + vt - 4.9t^2 and solve for h.
Given:
h = 150 meters (height of the tower)
v = 40 meters per second (initial velocity)
t = 2 seconds (time elapsed)
Substituting these values into the equation, we have:
h + vt - 4.9t^2 = 150 + 40(2) - 4.9(2)^2
First, we need to calculate the value of 2^2:
2^2 = 2 * 2 = 4
Now we can substitute this value into the equation:
h + vt - 4.9t^2 = 150 + 40(2) - 4.9(4)
Next, we compute each expression:
h + vt - 4.9t^2 = 150 + 80 - 19.6
Adding and subtracting the numbers, we get:
h + vt - 4.9t^2 = 230 - 19.6
Finally, we can calculate the last expression:
h + vt - 4.9t^2 = 210.4
Therefore, the height of the rocket after 2 seconds is approximately 210.4 meters (rounded to the nearest tenth).