A rocket starts from rest and moves upward from the surface of the earth. For the first 10s of its motion, the vertical acceleration of the rocket is given by ay=(3.00m/s3)t, where the +y-direction is upward. What is the height of the rocket above the surface of the earth at t = 10s ? What is the speed of the rocket when it is 350m above the surface of the earth? a) I tried s=1/2*at^2 = ((3.00 m/s^3)(10^2 s)) / 2 = 150m but that was not the correct answer. I am also not sure how the seconds cancel out (my equation really gives me m/s rather than just m). b) To find velocity at 350m distance above the earth, I have to find time so 350m = (1/2)(3.00 m/s^3)t^2 so t=15.28s Then v=at as the initial velocity is zero, so v = (3.00 m/s^3)(5.28s) = 45.84 m/s Can someone please let me know what I'm doing wrong?