A Rocket fired vertically burns its fuel for 25 seconds, which provides a thrust producing a constant upward acceleration of 8 m/s2. Determine (i)...

Denote the first 25 seconds as `t_1` and the acceleration during this time as `a.` Then the height is  `H_1(t) = (a t^2)/2` and the speed is  `V_1(t) = at.` So the final speed after fuel is burnt is  `v_1=a t_1`  and the final height is `h_1 = (a t_1^2)/2.`

The height after `t_1` is the usual free fall height:

`H_2(t) = h_1+v_1 (t-t_1) - (g (t-t_1)^2)/2.`

The maximum altitude is reached at  `t_2=t_1+v_1/g`  and it is  `h_2 = h_1+(v_1^2)/(2g) = a/2 t_1^2(1+a/g).`

Rocket falls when `H_2(t)=0,` `tgtt_1.` This is a quadratic equation and the solution is 

`t_3 = t_1+(v_1+sqrt(v_1^2+4*g/2*h_1))/(2*(g/2)) = t_1+(v_1+sqrt(v_1^2+2g h_1))/(g) =`

`= t_1+a/(g) t_1 (1+sqrt((a/g)^2+a/g)).`

The answers are: (i) `h_1,` (ii) `h_2,` (iii) `t_3.` In numbers they are (i) 2500 m, (ii) about 4541 m and (iii) about 70 s.