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Q. A rocket is fired upward from the earth 's surface such that it creates an acceleration of 19.6 ms^–2. If after 5 s, its engine is switched off, what would be the maximum height of the rocket from earth's surface ?

 

we have to calculate the maximum height from the earth surface  of the rocket after 5 second when initial acceleration is given. 

the initial acceleration  a =19.6 m/s^2 

acceleration due to gravity ,g = 9.8 m/s^2

the net acceleration get reduced due to gravity so it will become 

 a' = 19.6 -9.8  =9.8m/s^2

distance covered in 5 second  s = ut +1/2 at^2

                                                 =  0*5 + .5* 9.8 *25 = 122.5 m

velocity after 5 second  v = u+ at  = 0 + 9.8*5  =49 m/s

now after 5 second acceleration will be zero so net acceleration  = -9.8 m/s^2

now        s' = (v^2 - u^2)/2a 

                    = (0 -49^2)/-9.8*2 = 122.5 m

there fore the maximum height of rocket will be  = s + s'

                                                           = 122.5 + 122.5

   maximum height of rocket will be   =245 m