Answer:
solution:
when the engine are fired the rocket has a linear constant acceleration motion:
[tex]V_{1} ^{2} =v_{0} ^{2} +2a_{1} (y_{1} -y_{0} )t=(0)^2+2(16)(y_{1}-0)\\V_{1} ^{2}=32y_{1}................. eq(1)[/tex]
[tex]V_{1}[/tex] is the final velocity of the rocket
when the engines are fired it become equal to the initial velocity of the rocket,
when the engines are shut off
[tex]V_{2} ^{2} =v_{1} ^{2} +2a_{2} (y_{2} -y_{1} )t=>v_{1} ^{2}-2(9.8)(960-y_{1})\\V_{2} ^{2} =v_{1} ^{2}-18816+19.6y_{1}...................eq(2)[/tex]
solve eq(1) and eq(2) we find
[tex](0)^2=(32y_{1} )-18816+19.6y_{1}[/tex]
solving for [tex]y_{1}[/tex]=364.65 m
Where [tex]y_{1}[/tex] is the distance travelled by the rockets for shutting off the engine
when the engines are fired:
[tex]y_{1} =y_{o} + v_{0}t_{1} +\frac{1}{2}at^{2} =>0+(0)T+\frac{1}{2}(16)t^{2\\\\\\364.65=8T^{2} -->T=6.75s[/tex]
NOTE:
DIAGRAM IS ATTACHED
