In the above question, the acceleration of the rocket is :

Answer

20 m

Explanation

`u sin thetaxx1-1/2 g(1)^(2)=u sin thetaxx3-1/2xxgxx(3)^(2)`
`2u sin theta=40 rArr u sin theta=20 m//s`
Max. height `=(u^(2)sin^(2) theta)/(2g)=(20xx20)/2-=20 m`