so would the distance traveled to the max height during free fall acceleration be : 0= ((at^2/2)^2) +gy, and then y = -(a^2t^4)/4g?
and then would the maximum height be: (at^2)/2 - (a^2t^4)/4g ?
it doesn't look right, am i doing something wrong?
i've come to see that the initial velocity is 0, the velocity of the rocket when it runs out of fuel is a*t which i got from the equation v = v_0 + at.
I'm guessing the maximum height is the distance it travels from the ground to the point where it runs out of fuel plus the distance from...
A rocket on the ground, accelerates straight upward from rest with constant net acceleration "a" , until time "t" , when the fuel is depleted. Here "g" is a positive number equal to the magnitude of the acceleration "a" due to gravity.
-What is the maximum height reached in terms of a, t...