You have correctly calculated the first part of the question, assuming you are doing this question with simple algebra [SUVAT] it will work approximately as follows:
If M=25 and F=400, this means F=M*A, and A=16. With gravity NetA=16-9.81=6.19
We know D=U*T+0.5*A*(T^2), which means D=15*5+0.5*(6.19)*(5^2)=152.375m
This means Height above ground when the engine turns off should be 152.375+20=172.375m.
But you have to remember that after the 5 seconds (when the engine cuts off) the plane still keeps going, until gravity brings it to a stop and then turns it around:
We know FV=U+A*T=15+6.19*5=45.95, and with NetA=-9.81, you can now calculate the time taken to come to a stop as well as the extra distance it goes after the engine is turned off:
The time taken for it to stop is, T=(0-45.95)/-9.81=4.68sec
We know D=U*T+0.5*A*(T^2), which means D=45.95*4.68+0.5*(-9.81)*(4.68^2)=107.61m
So, the total distance it goes is TotalDist=172.375+107.61=279.985 [which due to rounding in the question has ended up being 280.5]
