A rocket takes of from a platform at 1m. After it takes off, its speed upward is 1/2 of the height that it has reached after t seconds. What is the equation for the height of the rocket after t seconds? Solution S=32+UT-GT^2/2...WHERE S=DISTANCE TRAVELLED INCLUDING THE INITIAL HT.OR TOTAL HEIGHT ABOVE GROUND......FT. U=INITIAL VELOCITY...FT/SEC T= TIME...SEC. G=ACCELERATION DUE TO GRAVITY...32 FT/SEC.SEC. SUBSTITUTING... S=32+154*T - 32*T^2/2 =32+154*T-16*T^2 i hope it helps you :)).