Question

During your summer internship for an aerospace company, you are asked to design a small research rocket. The rocket is to be launched from rest from the earth's surface and is to reach a maximum height of 940 m above the earth's surface. The rocket's engines give the rocket an upward acceleration so it moves with acceleration of 16.0 m/s2 during the time T that they fire. After the engines shut off, the rocket is in free fall. Ignore air resistance. Assume that the acceleration due to gravity does not change with the height of the rocket.

Answer 1

solution:

when the engine are fired the rocket has a linear constant acceleration motion:

`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)`

`V_(1)` 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

`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)`

solve eq(1) and eq(2) we find

`(0)^2=(32y_(1) )-18816+19.6y_(1)`

solving for
`y_(1)`=364.65 m

Where
`y_(1)` is the distance travelled by the rockets for shutting off the engine

when the engines are fired:

`y_(1) =y_(o) + v_(0)t_(1) +(1)/(2)at^(2) =>0+(0)T+(1)/(2)(16)t^{2\\\\\\364.65=8T^(2) -->T=6.75s`

NOTE:

DIAGRAM IS ATTACHED