rocket launched from a 90 meter high building, upward speed of 20 meters per second. what will it's height be after 3 seconds.
given the polynomial h + vt -4.9t^2
90 +(20) (3) - 4.9 (3)^2
Is the answer 130.4 ?
90 + 60 - 4.9(9) = 105.9
how did you get 130.4 ?
I will go back and check my numbers. Thanks for your help.
To solve this problem, we can use the given polynomial equation:
h + vt - 4.9t^2
where h is the initial height of the rocket (90 meters), v is the upward speed (20 meters per second), and t is the time (3 seconds).
Substituting the values into the equation, we have:
90 + (20)(3) - 4.9(3)^2
Calculating this expression, we get:
90 + 60 - 4.9(9)
= 90 + 60 - 44.1
= 150 - 44.1
= 105.9
Therefore, the height of the rocket after 3 seconds will be approximately 105.9 meters.
So, the answer is not 130.4.